The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle".

In this problem, you are supposed to tell if a given cycle is a Hamiltonian cycle.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers N (2< N <= 200), the number of vertices, and M, the number of edges in an undirected graph. Then M lines follow, each describes an edge in the format "Vertex1 Vertex2", where the vertices are numbered from 1 to N. The next line gives a positive integer K which is the number of queries, followed by K lines of queries, each in the format:

n V₁ V₂ ... V_n

where n is the number of vertices in the list, and V_i's are the vertices on a path.

Output Specification:

For each query, print in a line "YES" if the path does form a Hamiltonian cycle, or "NO" if not.

Sample Input:

6 10

6 2

3 4

1 5

2 5

3 1

4 1

1 6

6 3

1 2

4 5

6

7 5 1 4 3 6 2 5

6 5 1 4 3 6 2

9 6 2 1 6 3 4 5 2 6

4 1 2 5 1

7 6 1 3 4 5 2 6

7 6 1 2 5 4 3 1

Sample Output:

YES

NO

NO

NO

YES

NO


哈密顿圈 所有点只经过一次的回路。
代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int mp[][];

int n,m,a,b,c,v[],k;

int main()

{

    cin>>n>>m;

    for(int i = ;i < m;i ++)

    {

        cin>>a>>b;

        mp[a][b] = mp[b][a] = ;

    }

    cin>>m;

    for(int i = ;i < m;i ++)

    {

        cin>>k;

        int flag = ;

        if(k != n + )flag = ;

        cin>>a;

        c = a;

        memset(v,,sizeof(v));

        v[a] ++;

        for(int j = ;j < k;j ++)

        {

            cin>>b;

            if(!mp[a][b])flag = ;

            v[b] ++;

            if(v[b] ==  && b != c || v[b] > )flag = ;

            a = b;

        }

        if(b != c)flag = ;

        if(flag)cout<<"YES"<<endl;

        else cout<<"NO"<<endl;

    }

}