Codeforces Round #504 (rated, Div. 1 + Div. 2, based on VK Cup 2018 Final) ABCD

A. Single Wildcard Pattern Matching

You are given two strings $\mathrm{ss and tt.\; The\; string ss consists\; of\; lowercase\; Latin\; letters\; and at\; most\; one wildcard\; character\; \text{'}*\text{'},\; the\; string ttconsists\; only\; of\; lowercase\; Latin\; letters.\; The\; length\; of\; the\; string ss equals nn,\; the\; length\; of\; the\; string tt equals mm.}$

The wildcard character '*' in the string $\mathrm{ss (if\; any)\; can\; be\; replaced\; with\; an\; arbitrary\; sequence\; (possibly\; empty)\; of\; lowercase\; Latin\; letters.\; No\; other\; character\; of ss can\; be\; replaced\; with\; anything.\; If\; it\; is\; possible\; to\; replace\; a\; wildcard\; character\; \text{'}*\text{'}\; in ss to\; obtain\; a\; string tt,\; then\; the\; string tt matches\; the\; pattern ss.}$

For example, if $\mathrm{s=s="aba*aba"\; then\; the\; following\; strings\; match\; it\; "abaaba",\; "abacaba"\; and\; "abazzzaba",\; but\; the\; following\; strings\; do\; not\; match:\; "ababa",\; "abcaaba",\; "codeforces",\; "aba1aba",\; "aba?aba".}$

If the given string $\mathrm{tt matches\; the\; given\; string ss,\; print\; "YES",\; otherwise\; print\; "NO".}$

Input

The first line contains two integers $\mathrm{nn and mm (1\le n,m\le 2\cdot 1051\le n,m\le 2\cdot 105)\; —\; the\; length\; of\; the\; string ss and\; the\; length\; of\; the\; string tt,\; respectively.}$

The second line contains string $\mathrm{ss of\; length nn,\; which\; consists\; of\; lowercase\; Latin\; letters\; and at\; most\; one wildcard\; character\; \text{'}*\text{'}.}$

The third line contains string $\mathrm{tt of\; length mm,\; which\; consists\; only\; of\; lowercase\; Latin\; letters.}$

Output

Print "YES" (without quotes), if you can obtain the string $\mathrm{tt from\; the\; string ss.\; Otherwise\; print\; "NO"\; (without\; quotes).}$

Examples

input

Copy

6 10
code*s
codeforces

output

Copy

YES

input

Copy

6 5
vk*cup
vkcup

output

Copy

YES

input

Copy

1 1
v
k

output

Copy

NO

input

Copy

9 6
gfgf*gfgf
gfgfgf

output

Copy

NO

Note

In the first example a wildcard character '*' can be replaced with a string "force". So the string $\mathrm{ss after\; this\; replacement\; is\; "codeforces"\; and\; the\; answer\; is\; "YES".}$

In the second example a wildcard character '*' can be replaced with an empty string. So the string $\mathrm{ss after\; this\; replacement\; is\; "vkcup"\; and\; the\; answer\; is\; "YES".}$

There is no wildcard character '*' in the third example and the strings "v" and "k" are different so the answer is "NO".

In the fourth example there is no such replacement of a wildcard character '*' that you can obtain the string $\mathrm{tt so\; the\; answer\; is\; "NO".}$

 

$\mathrm{模拟题，有点费时。}$

#include <bits/stdc++.h>
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int N = 2e5+10;
char s[N], t[N];
int main() {
    int n, m, id;
    cin >> n >> m;
    cin >> s >> t;
    bool flag = false;
    for(int i = 0; i < n; i ++) {
        if(s[i] == '*') {
            flag = true;
            id = i;
            break;
        }
    }
    if(!flag) {
        if(n != m) return 0*printf("NO\n");
        for(int i = 0; i < n; i ++) {
            if(s[i] != t[i]) return 0*printf("NO\n");
        }
        printf("YES\n");
    } else {
        int l = 0, r = n-1, i = 0;
        while(s[l] == t[l]) l ++;
        while(s[r] == t[m-1-i]) {
            r--;i++;
        }
        // printf("%d %d %d\n",l,r,id);
        if(n <= m+1 && l == id && r == id) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

 

B. Pair of Toys

Tanechka is shopping in the toy shop. There are exactly $\mathrm{nn toys\; in\; the\; shop\; for\; sale,\; the\; cost\; of\; the ii-th\; toy\; is ii burles.\; She\; wants\; to\; choose\; two\; toys\; in\; such\; a\; way\; that\; their\; total\; cost\; is kk burles.\; How\; many\; ways\; to\; do\; that\; does\; she\; have?}$

Each toy appears in the shop exactly once. Pairs $(a,b)(a,b) and (b,a)(b,a) are\; considered\; equal.\; Pairs (a,b)(a,b),\; where a=ba=b,\; are\; not\; allowed.$

Input

The first line of the input contains two integers $\mathrm{nn, kk (1\le n,k\le 10141\le n,k\le 1014)\; —\; the\; number\; of\; toys\; and\; the\; expected\; total\; cost\; of\; the\; pair\; of\; toys.}$

Output

Print the number of ways to choose the pair of toys satisfying the condition above. Print 0, if Tanechka can choose no pair of toys in such a way that their total cost is $\mathrm{kk burles.}$

Examples

input

Copy

8 5

output

Copy

2

input

Copy

8 15

output

Copy

1

input

Copy

7 20

output

Copy

0

input

Copy

1000000000000 1000000000001

output

Copy

500000000000

Note

In the first example Tanechka can choose the pair of toys ($\mathrm{1,41,4)\; or\; the\; pair\; of\; toys\; (2,32,3).}$

In the second example Tanechka can choose only the pair of toys ($\mathrm{7,87,8).}$

In the third example choosing any pair of toys will lead to the total cost less than $\mathrm{2020.\; So\; the\; answer\; is 0.}$

In the fourth example she can choose the following pairs: $(1,1000000000000)(1,1000000000000), (2,999999999999)(2,999999999999), (3,999999999998)(3,999999999998),\; ..., (500000000000,500000000001)(500000000000,500000000001).\; The\; number\; of\; such\; pairs\; is\; exactly 500000000000500000000000.$

$数学题$

#include <bits/stdc++.h>
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int N = 110;

int main() {
    ll n, k;
    cin >> n >> k;
    printf("%lld\n",min(k-1,max(k/2,n))-k/2);
    return 0;
}

 

C. Bracket Subsequence

A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not.

Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

You are given a regular bracket sequence $\mathrm{ss and\; an\; integer\; number kk.\; Your\; task\; is\; to\; find\; a\; regular\; bracket\; sequence\; of\; length\; exactly kksuch\; that\; it\; is\; also\; a\; subsequence\; of ss.}$

It is guaranteed that such sequence always exists.

Input

The first line contains two integers $\mathrm{nn and kk (2\le k\le n\le 2\cdot 1052\le k\le n\le 2\cdot 105,\; both nn and kk are\; even)\; —\; the\; length\; of ss and\; the\; length\; of\; the\; sequence\; you\; are\; asked\; to\; find.}$

The second line is a string $\mathrm{ss —\; regular\; bracket\; sequence\; of\; length nn.}$

Output

Print a single string — a regular bracket sequence of length exactly $\mathrm{kk such\; that\; it\; is\; also\; a\; subsequence\; of ss.}$

It is guaranteed that such sequence always exists.

Examples

input

Copy

6 4
()(())

output

Copy

()()

input

Copy

8 8
(()(()))

output

Copy

(()(()))

去掉相对应的括号

#include <bits/stdc++.h>
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int N = 2e5+10;
char s[N];
int a[N], vis[N];
int main() {
       int n, k;
       cin >> n >> k;
       cin >> s+1;
       stack<int> st;
       for(int i = 1; i <= n; i ++) {
           if(s[i] == '(') st.push(i);
           else {
               int cnt = st.top();
               st.pop();
               a[cnt] = i;
               a[i] = cnt;
           }
       }
       int l = 1;
       for(int i = 0; i < (n-k)/2; i ++) {
           while(vis[l])l++;
          vis[l] = vis[a[l]] = 1;
       }
       for(int i = 1; i <= n; i ++) {
           if(!vis[i])printf("%c",s[i]);
       }printf("\n");
    return 0;
}

 

D. Array Restoration

Initially there was an array $\mathrm{aa consisting\; of nn integers.\; Positions\; in\; it\; are\; numbered\; from 11 to nn.}$

Exactly $\mathrm{qq queries\; were\; performed\; on\; the\; array.\; During\; the ii-th\; query\; some\; segment (li,ri)(li,ri) (1\le li\le ri\le n)(1\le li\le ri\le n) was\; selected\; and\; values\; of\; elements\; on\; positions\; from lili to riri inclusive\; got\; changed\; to ii.\; The\; order\; of\; the\; queries\; couldn\text{'}t\; be\; changed\; and\; all qq queries\; were\; applied.\; It\; is\; also\; known\; that\; every\; position\; from 11 to nn got\; covered\; by\; at\; least\; one\; segment.}$

We could have offered you the problem about checking if some given array (consisting of $\mathrm{nn integers\; with\; values\; from 11 to qq)\; can\; be\; obtained\; by\; the\; aforementioned\; queries.\; However,\; we\; decided\; that\; it\; will\; come\; too\; easy\; for\; you.}$

So the enhancement we introduced to it is the following. Some set of positions (possibly empty) in this array is selected and values of elements on these positions are set to $00.$

Your task is to check if this array can be obtained by the aforementioned queries. Also if it can be obtained then restore this array.

If there are multiple possible arrays then print any of them.

Input

The first line contains two integers $\mathrm{nn and qq (1\le n,q\le 2\cdot 1051\le n,q\le 2\cdot 105)\; —\; the\; number\; of\; elements\; of\; the\; array\; and\; the\; number\; of\; queries\; perfomed\; on\; it.}$

The second line contains $\mathrm{nn integer\; numbers a1,a2,\dots ,ana1,a2,\dots ,an (0\le ai\le q0\le ai\le q)\; —\; the\; resulting\; array.\; If\; element\; at\; some\; position jj is\; equal\; to 00then\; the\; value\; of\; element\; at\; this\; position\; can\; be\; any\; integer\; from 11 to qq.}$

Output

Print "YES" if the array $\mathrm{aa can\; be\; obtained\; by\; performing qq queries.\; Segments (li,ri)(li,ri) (1\le li\le ri\le n)(1\le li\le ri\le n) are\; chosen\; separately\; for\; each\; query.\; Every\; position\; from 11 to nn should\; be\; covered\; by\; at\; least\; one\; segment.}$

Otherwise print "NO".

If some array can be obtained then print $\mathrm{nn integers\; on\; the\; second\; line\; —\; the ii-th\; number\; should\; be\; equal\; to\; the ii-th\; element\; of\; the\; resulting\; array\; and\; should\; have\; value\; from 11 to qq.\; This\; array\; should\; be\; obtainable\; by\; performing\; exactly qq queries.}$

If there are multiple possible arrays then print any of them.

Examples

input

Copy

4 3
1 0 2 3

output

Copy

YES
1 2 2 3

input

Copy

3 10
10 10 10

output

Copy

YES
10 10 10 

input

Copy

5 6
6 5 6 2 2

output

Copy

NO

input

Copy

3 5
0 0 0

output

Copy

YES
5 4 2

Note

In the first example you can also replace $\mathrm{00 with 11 but\; not\; with 33.}$

In the second example it doesn't really matter what segments to choose until query $\mathrm{1010 when\; the\; segment\; is (1,3)(1,3).}$

The third example showcases the fact that the order of queries can't be changed, you can't firstly set $(1,3)(1,3) to 66 and\; after\; that\; change (2,2)(2,2) to 55.\; The\; segment\; of 55 should\; be\; applied\; before\; segment\; of 66.$

There is a lot of correct resulting arrays for the fourth example.

 

0.0  

#include <bits/stdc++.h>
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int N = 2e5+10;
int a[N], n, q, ans[N];
int dp[2][N];
int main() {
    scanf("%d%d",&n, &q);
    memset(dp[0], INF, sizeof(dp[0]));
    memset(dp[1], 0, sizeof(dp[1]));
    for(int i = 1; i <= n; i ++) {
        scanf("%d",&a[i]);
        dp[0][a[i]] = min(dp[0][a[i]], i);
        dp[1][a[i]] = max(dp[1][a[i]], i);
    }
    for(int i = q; i >= 1; i --) {
        int l = dp[0][i], r = dp[1][i];
        if(r == 0) continue;
        for(int j = l,k = r;j <= k; j ++, k --) {
            if(ans[j] && ans[k]) break;
            if(a[j] == 0) a[j] = i;
            if(a[k] == 0) a[k] = i;
            if(a[j] == i) ans[j] = 1;
            else if(a[j] < i) return 0*printf("NO\n");
            if(a[k] == i) ans[k] = 1;
            else if(a[k] < i) return 0*printf("NO\n");
        }
    }
    if(dp[1][q] == 0) {
        if(dp[1][0] == 0) return 0*printf("NO\n");
        else {
            a[dp[0][0]]=q;
            for(int i = 1;i <= n; i ++)  if(a[i]==0) a[i]=a[i-1];
            printf("YES\n");
            for(int i = 1;i <= n; i ++) printf("%d%c",a[i]," \n"[i==n]);
        }
    } else {
        int cnt = 0;
        for(int i = 1; i <= n; i ++)
            if(a[i]){
                cnt = i;
                break;
            }
        for(int i = cnt-1, j = cnt+1; i >= 1 || j <= n; i --, j ++) {
            if(i >= 1 && a[i] == 0) a[i]=a[i+1];
            if(j <= n && a[j] == 0) a[j]=a[j-1];
        }
        printf("YES\n");
        for(int i = 1;i <= n; i ++) printf("%d%c",a[i]," \n"[i==n]);
    }
    return 0;
}