Codeforces Round #473 (Div. 2)

 

A. Mahmoud and Ehab and the even-odd game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:

• 1 ≤ a ≤ n.
• If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd.

If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?

Input

The only line contains an integer n (1 ≤ n ≤ 109), the number at the beginning of the game.

Output

Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.

Examples

input

Copy

1

output

Ehab

input

Copy

2

output

Mahmoud

Note

In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.

In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.

A就是Ehab要拿奇数的，Mahmoud要拿偶数个，直接拿到0啊

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    if(n&1)cout<<"Ehab";
    else cout<<"Mahmoud";
    return 0;
}

B. Mahmoud and Ehab and the message

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Mahmoud wants to send a message to his friend Ehab. Their language consists of n words numbered from 1 to n. Some words have the same meaning so there are k groups of words such that all the words in some group have the same meaning.

Mahmoud knows that the i-th word can be sent with cost ai. For each word in his message, Mahmoud can either replace it with another word of the same meaning or leave it as it is. Can you help Mahmoud determine the minimum cost of sending the message?

The cost of sending the message is the sum of the costs of sending every word in it.

Input

The first line of input contains integers n, k and m (1 ≤ k ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of words in their language, the number of groups of words, and the number of words in Mahmoud's message respectively.

The second line contains n strings consisting of lowercase English letters of length not exceeding 20 which represent the words. It's guaranteed that the words are distinct.

The third line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) where ai is the cost of sending the i-th word.

The next k lines describe the groups of words of same meaning. The next k lines each start with an integer x (1 ≤ x ≤ n) which means that there are x words in this group, followed by x integers which represent the indices of words in this group. It's guaranteed that each word appears in exactly one group.

The next line contains m space-separated words which represent Mahmoud's message. Each of these words appears in the list of language's words.

Output

The only line should contain the minimum cost to send the message after replacing some words (maybe none) with some words of the same meaning.

Examples

input

Copy

5 4 4
i loser am the second
100 1 1 5 10
1 1
1 3
2 2 5
1 4
i am the second

output

107

input

Copy

5 4 4
i loser am the second
100 20 1 5 10
1 1
1 3
2 2 5
1 4
i am the second

output

116

Note

In the first sample, Mahmoud should replace the word "second" with the word "loser" because it has less cost so the cost will be 100+1+5+1=107.

In the second sample, Mahmoud shouldn't do any replacement so the cost will be 100+1+5+10=116.

求最小，做一下hash映射

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
typedef long long ll;
const ll INF=0x3f3f3f3f3f3f3f3f;
string s[N];
ll a[N],b[N];
map<string,ll> M;
int main()
{
    ios::sync_with_stdio(false);
    int n,k,m,x;
    ll ans=0;
    cin>>n>>k>>m;
    for(int i=1; i<=n; i++)
        cin>>s[i];
    for(int i=1; i<=n; i++)cin>>a[i];
    for(int i=1,x; i<=k; i++)
    {
        cin>>x;
        ll mi=INF;
        for(int j=1; j<=x; j++)
        {
            cin>>b[j];
            mi=min(mi,a[b[j]]);
        }
        for(int j=1; j<=x; j++)
            M[s[b[j]]]=mi;
    }
    string c;
    for(int i=1; i<=m; i++)
        cin>>c,ans+=M[c];
    cout<<ans;
    return 0;
}

C. Mahmoud and Ehab and the wrong algorithm

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Mahmoud was trying to solve the vertex cover problem on trees. The problem statement is:

Given an undirected tree consisting of n nodes, find the minimum number of vertices that cover all the edges. Formally, we need to find a set of vertices such that for each edge (u, v) that belongs to the tree, either u is in the set, or v is in the set, or both are in the set. Mahmoud has found the following algorithm:

• Root the tree at node 1.
• Count the number of nodes at an even depth. Let it be evenCnt.
• Count the number of nodes at an odd depth. Let it be oddCnt.
• The answer is the minimum between evenCnt and oddCnt.

The depth of a node in a tree is the number of edges in the shortest path between this node and the root. The depth of the root is 0.

Ehab told Mahmoud that this algorithm is wrong, but he didn't believe because he had tested his algorithm against many trees and it worked, so Ehab asked you to find 2 trees consisting of n nodes. The algorithm should find an incorrect answer for the first tree and a correct answer for the second one.

Input

The only line contains an integer n (2 ≤ n ≤ 105), the number of nodes in the desired trees.

Output

The output should consist of 2 independent sections, each containing a tree. The algorithm should find an incorrect answer for the tree in the first section and a correct answer for the tree in the second. If a tree doesn't exist for some section, output "-1" (without quotes) for that section only.

If the answer for a section exists, it should contain n - 1 lines, each containing 2 space-separated integers u and v (1 ≤ u, v ≤ n), which means that there's an undirected edge between node u and node v. If the given graph isn't a tree or it doesn't follow the format, you'll receive wrong answer verdict.

If there are multiple answers, you can print any of them.

Examples

input

Copy

2

output

-1
1 2

input

Copy

8

output

1 2
1 3
2 4
2 5
3 6
4 7
4 8
1 2
1 3
2 4
2 5
2 6
3 7
6 8

Note

In the first sample, there is only 1 tree with 2 nodes (node 1 connected to node 2). The algorithm will produce a correct answer in it so we printed  - 1 in the first section, but notice that we printed this tree in the second section.

In the second sample:

In the first tree, the algorithm will find an answer with 4 nodes, while there exists an answer with 3 nodes like this:In the second tree, the algorithm will find an answer with 3 nodes which is correct:

这个构造还是十分巧妙的

#include<bits/stdc++.h>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    if(n<6)cout<<-1<<"\n";
    else
    {
        cout<<"1 2\n";
        cout<<"2 3\n";
        cout<<"2 4\n";
        cout<<"3 5\n";
        cout<<"3 6\n";
        for(int i=7; i<=n; i++)cout<<3<<" "<<i<<"\n";
    }
    for(int i=2; i<=n; i++)cout<<i-1<<" "<<i<<"\n";
    return 0;
}

E. Mahmoud and Ehab and the xor-MST

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight  (where  is the bitwise-xor operation). Can you find the weight of the minimum spanning tree of that graph?

You can read about complete graphs in https://en.wikipedia.org/wiki/Complete_graph

You can read about the minimum spanning tree in https://en.wikipedia.org/wiki/Minimum_spanning_tree

The weight of the minimum spanning tree is the sum of the weights on the edges included in it.

Input

The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.

Output

The only line contains an integer x, the weight of the graph's minimum spanning tree.

Example

input

Copy

4

output

4

Note

In the first sample:The weight of the minimum spanning tree is 1+2+1=4.

这些都不用考虑，一顿操作猛如虎，先上打表代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x7fffffff
//Prim算法实现
void la(int n)
{
    char f[100];
   static char c[16]= {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
    int l=0;
    while(n) {
        f[l++]=c[n%2];
        n=n/2;
    }
    for(int k=0; k<l; k++)
    printf("%c ",f[k]);
    cout<<"\n";
}
void prim_test()
{
    freopen("out.txt","w",stdout);
    for(int n=1; n<=500; n++)
    {
        vector<vector<int> > A(n, vector<int>(n));
        for(int k=0; k<n; k++)
            for(int l=k+1; l<n; l++)
            {
                int cost=l^k;
                A[k][l] = cost;
                A[l][k] = cost;
            }

        int pos, minimum;
        int min_tree = 0;
        vector<int> visited, lowcost;
        for (int i = 0; i < n; ++i)
        {
            visited.push_back(0);    //初始化为0，表示都没加入
        }
        visited[0] = 1;   //最小生成树从第一个顶点开始
        for (int i = 0; i < n; ++i)
        {
            lowcost.push_back(A[0][i]);    //权值初始化为0
        }

        for (int i = 0; i < n; ++i)      //枚举n个顶点
        {
            minimum = INF;
            for (int j = 0; j < n; ++j)      //找到最小权边对应顶点
            {
                if(!visited[j] && minimum > lowcost[j])
                {
                    minimum = lowcost[j];
                    pos = j;
                }
            }
            if (minimum == INF)    //如果min = INF表示已经不再有点可以加入最小生成树中
                break;
            min_tree += minimum;
            visited[pos] = 1;     //加入最小生成树中
            for (int j = 0; j < n; ++j)
            {
                if(!visited[j] && lowcost[j] > A[pos][j]) lowcost[j] = A[pos][j];   //更新可更新边的权值
            }
        }
        la(min_tree);
    }
}

int main(void)
{
    prim_test();

    return 0;
}

数字还是十分有规律的，其实就是统计2进制1的个数，但是自己太菜了，竟然没有做出来

#include <bits/stdc++.h>
using namespace std;
long long la(long long m)
{
    if(m==0)return 0;
    return 2LL*la(m/2)+(m+1)/2;
}
int main()
{
    long long n;
    cin>>n;
    cout<<la(n-1);
    return 0;
}

D. Mahmoud and Ehab and another array construction task

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Mahmoud has an array a consisting of n integers. He asked Ehab to find another array b of the same length such that:

• b is lexicographically greater than or equal to a.
• bi ≥ 2.
• b is pairwise coprime: for every 1 ≤ i < j ≤ n, bi and bj are coprime, i. e. GCD(bi, bj) = 1, where GCD(w, z) is the greatest common divisor of w and z.

Ehab wants to choose a special array so he wants the lexicographically minimal array between all the variants. Can you find it?

An array x is lexicographically greater than an array y if there exists an index i such than xi > yi and xj = yj for all 1 ≤ j < i. An array x is equal to an array y if xi = yi for all 1 ≤ i ≤ n.

Input

The first line contains an integer n (1 ≤ n ≤ 105), the number of elements in a and b.

The second line contains n integers a1, a2, ..., an (2 ≤ ai ≤ 105), the elements of a.

Output

Output n space-separated integers, the i-th of them representing bi.

Examples

input

Copy

5
2 3 5 4 13

output

2 3 5 7 11 

input

Copy

3
10 3 7

output

10 3 7 

Note

Note that in the second sample, the array is already pairwise coprime so we printed it.

预处理素数，然后模拟

#include <bits/stdc++.h>
using namespace std;
const int N=2000005;
vector<int> d[N];
int p[N],vis[N];
int ok(int x)
{
    int ret=1;
    for(auto i:d[x])
    ret&=!vis[i];
    return ret;
}
int main()
{
    int n;
    cin>>n;
    for(int i=2;i<N;i++)
        if (!p[i])
        {
            for(int x=i;x<N;x+=i)d[x].push_back(i),p[x]=1;
        }
    int f=1,cur=2;
    for(int i=0,x;i<n;i++)
    {
        cin>>x;
        int out=x;
        if(f)
        {
            while(!ok(out))
            out++;
            if(out!=x)f=0;
        }
        else
        {
            while(!ok(cur))cur++;
            out=cur;
        }
        printf("%d ",out);
        for(auto x:d[out])
        vis[x]=1;
    }
}