Manthan, Codefest 18 (rated, Div. 1 + Div. 2) ABCDE

A. Packets

You have $\mathrm{nn coins,\; each\; of\; the\; same\; value\; of 11.}$

Distribute them into packets such that any amount $\mathrm{xx (1\le x\le n1\le x\le n)\; can\; be\; formed\; using\; some\; (possibly\; one\; or\; all)\; number\; of\; these\; packets.}$

Each packet may only be used entirely or not used at all. No packet may be used more than once in the formation of the single $\mathrm{xx,\; however\; it\; may\; be\; reused\; for\; the\; formation\; of\; other xx\text{'}s.}$

Find the minimum number of packets in such a distribution.

Input

The only line contains a single integer $\mathrm{nn (1\le n\le 1091\le n\le 109) —\; the\; number\; of\; coins\; you\; have.}$

Output

Output a single integer — the minimum possible number of packets, satisfying the condition above.

Examples

input

Copy

6

output

Copy

3

input

Copy

2

output

Copy

2

Note

In the first example, three packets with $\mathrm{11, 22 and 33 coins\; can\; be\; made\; to\; get\; any\; amount xx (1\le x\le 61\le x\le 6).}$

• To get $\mathrm{11 use\; the\; packet\; with 11 coin.}$
• To get $\mathrm{22 use\; the\; packet\; with 22 coins.}$
• To get $\mathrm{33 use\; the\; packet\; with 33 coins.}$
• To get $\mathrm{44 use\; packets\; with 11 and 33 coins.}$
• To get $\mathrm{55 use\; packets\; with 22 and 33 coins}$
• To get $\mathrm{66 use\; all\; packets.}$

In the second example, two packets with $\mathrm{11 and 11 coins\; can\; be\; made\; to\; get\; any\; amount xx (1\le x\le 21\le x\le 2).}$

将n最少分成多少个数，使的1-n 都可以由这些数组成。  看成二进制，分成1 10 100 1000...就行。

#include <bits/stdc++.h>
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

int main() {
    int n;
    cin >> n;
    printf("%d\n",(int)log2(n)+1);
    return 0;
}

B. Reach Median

You are given an array $\mathrm{aa of nn integers\; and\; an\; integer ss.\; It\; is\; guaranteed\; that nn is\; odd.}$

In one operation you can either increase or decrease any single element by one. Calculate the minimum number of operations required to make the median of the array being equal to $\mathrm{ss.}$

The median of the array with odd length is the value of the element which is located on the middle position after the array is sorted. For example, the median of the array $\mathrm{6,5,86,5,8 is\; equal\; to 66,\; since\; if\; we\; sort\; this\; array\; we\; will\; get 5,6,85,6,8,\; and 66 is\; located\; on\; the\; middle\; position.}$

Input

The first line contains two integers $\mathrm{nn and ss (1\le n\le 2\cdot 105-11\le n\le 2\cdot 105-1, 1\le s\le 1091\le s\le 109) —\; the\; length\; of\; the\; array\; and\; the\; required\; value\; of\; median.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109) —\; the\; elements\; of\; the\; array aa.}$

It is guaranteed that $\mathrm{nn is\; odd.}$

Output

In a single line output the minimum number of operations to make the median being equal to $\mathrm{ss.}$

Examples

input

Copy

3 8
6 5 8

output

Copy

2

input

Copy

7 20
21 15 12 11 20 19 12

output

Copy

6

Note

In the first sample, $\mathrm{66 can\; be\; increased\; twice.\; The\; array\; will\; transform\; to 8,5,88,5,8,\; which\; becomes 5,8,85,8,8 after\; sorting,\; hence\; the\; median\; is\; equal\; to 88.}$

In the second sample, $\mathrm{1919 can\; be\; increased\; once\; and 1515 can\; be\; increased\; five\; times.\; The\; array\; will\; become\; equal\; to 21,20,12,11,20,20,1221,20,12,11,20,20,12.\; If\; we\; sort\; this\; array\; we\; get 11,12,12,20,20,20,2111,12,12,20,20,20,21,\; this\; way\; the\; median\; is 2020.}$

将中位数变成s，排序后看最中间的数是多少，然后调整其它数字为s。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5+10;
ll a[N];
int main() {
    ll n, s;
    cin >> n >> s;
    for(int i = 1; i <= n; i ++) cin >> a[i];
    sort(a+1,a+1+n);
    ll ans = 0;
    if(a[n/2+1] < s) {
        for(int i = n/2+1; i <= n; i ++) {
            if(a[i] < s) ans += s-a[i];
        }
    } else {
        for(int i = n/2+1; i >= 1; -- i) {
            if(a[i] > s) ans += a[i] - s;
        }
    }
    printf("%lld\n",ans);
    return 0;
}

C. Equalize

You are given two binary strings $\mathrm{aa and bb of\; the\; same\; length.\; You\; can\; perform\; the\; following\; two\; operations\; on\; the\; string aa:}$

• Swap any two bits at indices $\mathrm{ii and jj respectively\; (1\le i,j\le n1\le i,j\le n),\; the\; cost\; of\; this\; operation\; is |i-j||i-j|,\; that\; is,\; the\; absolute\; difference\; between ii and jj.}$
• Select any arbitrary index $\mathrm{ii (1\le i\le n1\le i\le n)\; and\; flip\; (change 00 to 11 or 11 to 00)\; the\; bit\; at\; this\; index.\; The\; cost\; of\; this\; operation\; is 11.}$

Find the minimum cost to make the string $\mathrm{aa equal\; to bb.\; It\; is\; not\; allowed\; to\; modify\; string bb.}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1061\le n\le 106) —\; the\; length\; of\; the\; strings aa and bb.}$

The second and third lines contain strings $\mathrm{aa and bb respectively.}$

Both strings $\mathrm{aa and bb have\; length nn and\; contain\; only\; \text{'}0\text{'}\; and\; \text{'}1\text{'}.}$

Output

Output the minimum cost to make the string $\mathrm{aa equal\; to bb.}$

Examples

input

Copy

3
100
001

output

Copy

2

input

Copy

4
0101
0011

output

Copy

1

Note

In the first example, one of the optimal solutions is to flip index $\mathrm{11 and\; index 33,\; the\; string aa changes\; in\; the\; following\; way:\; "100" \to \to "000" \to \to "001".\; The\; cost\; is 1+1=21+1=2.}$

The other optimal solution is to swap bits and indices $\mathrm{11 and 33,\; the\; string aa changes\; then\; "100" \to \to "001",\; the\; cost\; is\; also |1-3|=2|1-3|=2.}$

In the second example, the optimal solution is to swap bits at indices $\mathrm{22 and 33,\; the\; string aa changes\; as\; "0101" \to \to "0011".\; The\; cost\; is |2-3|=1|2-3|=1.}$

只有当连续两个位置都不同，且同一字符串中也不同是选择交换更好，其它都直接改变数字。

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6+10;
char s1[N], s2[N];
int main() {
    int n;
    cin >> n >> s1 >> s2;
    int ans = 0;
    for(int i = 0; i < n; i ++) {
        if(s1[i] != s2[i]) {
            if(i+1 < n && s1[i+1] != s2[i+1] && s1[i] != s1[i+1]) {
                s1[i+1] = s2[i+1];
            }
            ans++;
        }
    }
    printf("%d\n",ans);
    return 0;
}

D. Valid BFS?

The BFS algorithm is defined as follows.

1. Consider an undirected graph with vertices numbered from $\mathrm{11 to nn.\; Initialize qq as\; a\; new queue containing\; only\; vertex 11,\; mark\; the\; vertex 11 as\; used.}$
2. Extract a vertex $\mathrm{vv from\; the\; head\; of\; the\; queue qq.}$
3. Print the index of vertex $\mathrm{vv.}$
4. Iterate in arbitrary order through all such vertices $\mathrm{uu that uu is\; a\; neighbor\; of vv and\; is\; not\; marked\; yet\; as\; used.\; Mark\; the\; vertex uu as\; used\; and\; insert\; it\; into\; the\; tail\; of\; the\; queue qq.}$
5. If the queue is not empty, continue from step 2.
6. Otherwise finish.

Since the order of choosing neighbors of each vertex can vary, it turns out that there may be multiple sequences which BFS can print.

In this problem you need to check whether a given sequence corresponds to some valid BFS traversal of the given tree starting from vertex $\mathrm{11.\; The tree is\; an\; undirected\; graph,\; such\; that\; there\; is\; exactly\; one\; simple\; path\; between\; any\; two\; vertices.}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 2\cdot 1051\le n\le 2\cdot 105)\; which\; denotes\; the\; number\; of\; nodes\; in\; the\; tree.}$

The following $\mathrm{n-1n-1 lines\; describe\; the\; edges\; of\; the\; tree.\; Each\; of\; them\; contains\; two\; integers xx and yy (1\le x,y\le n1\le x,y\le n) —\; the\; endpoints\; of\; the\; corresponding\; edge\; of\; the\; tree.\; It\; is\; guaranteed\; that\; the\; given\; graph\; is\; a\; tree.}$

The last line contains $\mathrm{nn distinct\; integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le n1\le ai\le n) —\; the\; sequence\; to\; check.}$

Output

Print "Yes" (quotes for clarity) if the sequence corresponds to some valid BFS traversal of the given tree and "No" (quotes for clarity) otherwise.

You can print each letter in any case (upper or lower).

Examples

input

Copy

4
1 2
1 3
2 4
1 2 3 4

output

Copy

Yes

input

Copy

4
1 2
1 3
2 4
1 2 4 3

output

Copy

No

Note

Both sample tests have the same tree in them.

In this tree, there are two valid BFS orderings:

• $\mathrm{1,2,3,41,2,3,4,}$
• $\mathrm{1,3,2,41,3,2,4.}$

The ordering $\mathrm{1,2,4,31,2,4,3 doesn\text{'}t\; correspond\; to\; any\; valid BFS order.}$

先根据已知的树，计算每个点的层数，然后根据给定的序列，看能不能bfs输出。

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
vector<int> vs[N];
map<int,int> mp[N];
int a[N], d[N];
int main() {
    int n, x, y;
    cin >> n;
    for(int i = 1; i < n; i ++) {
        cin >> x >> y;
        vs[x].push_back(y);
        vs[y].push_back(x);
        mp[x][y] = mp[y][x] = 1;
    }
    for(int i = 1; i <= n; i ++) cin >> a[i];
    d[1] = 1;
    queue<int> que;
    que.push(1);
    while(que.size()) {
        int v = que.front(); que.pop();
        for(auto u : vs[v]) {
            if(d[u] == 0) {
                d[u] = d[v] + 1;
                que.push(u);
            }
        }
    }
    // for(int i = 1; i <= n; i ++) printf("%d ",d[i]);printf("\n");
    for(int i = 1; i < n; i ++) {
        if(d[a[i]] > d[a[i+1]]) return 0*printf("No\n");
    }
    int l = 2;
    que.push(1);
    while(que.size()) {
        int v = que.front(); que.pop();
        while(l <= n) {
            if(mp[v].count(a[l])) {
                que.push(a[l]); l ++;
            } else break;
        }
    }
    if(l == n+1) printf("Yes\n");
    else printf("No\n");
    return 0;
}

E. Trips

There are $\mathrm{nn persons\; who\; initially\; don\text{'}t\; know\; each\; other.\; On\; each\; morning,\; two\; of\; them,\; who\; were\; not\; friends\; before,\; become\; friends.}$

We want to plan a trip for every evening of $\mathrm{mm days.\; On\; each\; trip,\; you\; have\; to\; select\; a\; group\; of\; people\; that\; will\; go\; on\; the\; trip.\; For\; every\; person,\; one\; of\; the\; following\; should\; hold:}$

• Either this person does not go on the trip,
• Or at least $\mathrm{kk of\; his\; friends\; also\; go\; on\; the\; trip.}$

Note that the friendship is not transitive. That is, if $\mathrm{aa and bb are\; friends\; and bb and cc are\; friends,\; it\; does\; not\; necessarily\; imply\; that aa and cc are\; friends.}$

For each day, find the maximum number of people that can go on the trip on that day.

Input

The first line contains three integers $\mathrm{nn, mm,\; and kk (2\le n\le 2\cdot 105,1\le m\le 2\cdot 1052\le n\le 2\cdot 105,1\le m\le 2\cdot 105, 1\le k<n1\le k<n) —\; the\; number\; of\; people,\; the\; number\; of\; days\; and\; the\; number\; of\; friends\; each\; person\; on\; the\; trip\; should\; have\; in\; the\; group.}$

The $\mathrm{ii-th\; (1\le i\le m1\le i\le m)\; of\; the\; next mm lines\; contains\; two\; integers xx and yy (1\le x,y\le n1\le x,y\le n, x\ne yx\ne y),\; meaning\; that\; persons xx and yy become\; friends\; on\; the\; morning\; of\; day ii.\; It\; is\; guaranteed\; that xx and yy were\; not\; friends\; before.}$

Output

Print exactly $\mathrm{mm lines,\; where\; the ii-th\; of\; them\; (1\le i\le m1\le i\le m)\; contains\; the\; maximum\; number\; of\; people\; that\; can\; go\; on\; the\; trip\; on\; the\; evening\; of\; the\; day ii.}$

Examples

input

Copy

4 4 2
2 3
1 2
1 3
1 4

output

Copy

0
0
3
3

input

Copy

5 8 2
2 1
4 2
5 4
5 2
4 3
5 1
4 1
3 2

output

Copy

0
0
0
3
3
4
4
5

input

Copy

5 7 2
1 5
3 2
2 5
3 4
1 2
5 3
1 3

output

Copy

0
0
0
0
3
4
4

Note

In the first example,

• $\mathrm{1,2,31,2,3 can\; go\; on\; day 33 and 44.}$

In the second example,

• $\mathrm{2,4,52,4,5 can\; go\; on\; day 44 and 55.}$
• $\mathrm{1,2,4,51,2,4,5 can\; go\; on\; day 66 and 77.}$
• $\mathrm{1,2,3,4,51,2,3,4,5 can\; go\; on\; day 88.}$

In the third example,

• $\mathrm{1,2,51,2,5 can\; go\; on\; day 55.}$
• $\mathrm{1,2,3,51,2,3,5 can\; go\; on\; day 66 and 77.}$

这解题思想真巧妙，离线处理，每拆除一堆朋友关系，然后把朋友小于k的点去除，就可以得到答案。

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
set<int> st[N], s;
int ans[N], x[N], y[N];
int n, m, k;
void dfs(int v) {
    if(st[v].size() >= k || !s.count(v)) return;
    s.erase(v);
    for(auto u:st[v]) {
        st[u].erase(v);
        if(s.count(u) && st[u].size() < k) {
            dfs(u);
        }
    }
}

int main() {
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 1; i <= m; i ++) {
        scanf("%d%d", &x[i], &y[i]);
        st[x[i]].insert(y[i]);
        st[y[i]].insert(x[i]);
    }
    for(int i = 1; i <= n; ++ i) s.insert(i);
    for(int i = 1; i <= n; i ++) {
        if(st[i].size() < k) {
            dfs(i);
        }
    }
    ans[m] = s.size();
    for(int i = m; i > 1; i --) {
        st[x[i]].erase(y[i]);
        st[y[i]].erase(x[i]);
        if(st[x[i]].size() < k) dfs(x[i]);
        if(st[y[i]].size() < k) dfs(y[i]);
        ans[i-1] = s.size();
    }
    for(int i = 1; i <= m; i ++) printf("%d\n",ans[i]);
    return 0;
}