#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <assert.h>
#include <memory.h>
#include <queue>
using namespace std;

#define N 200002

vector<int >edge[N];

char s[N];
int vis[N][4];
char t[4] ={'A','A','B','B'};
bool dfs(int node,int ord) {
  if(vis[node][ord]==1) {
    return true;
  }
  vis[node][ord] = 1;
  char tt = t[(ord+1)%4];
  for(int i=0;i<edge[node].size();++i) {
    int ch = edge[node][i];
    char g = s[ch];
    if(tt==g) {
      if(dfs(ch,(ord+1)%4)) {
        return true;
      }
    }
  }
  vis[node][ord] = 2;
  return false;
}
void solve() {
  int n,m;scanf("%d %d",&n,&m);
  scanf(" %s",s+1);
  for(int i=m;i;--i) {
    int a,b;scanf("%d %d",&a,&b);
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  for(int i=1;i<=n;++i) {
    if(!vis[i][0] && s[i]=='A') {
      if(dfs(i,0)) {
        printf("Yes\n" );
        return;
      }
    }
  }
  printf("No\n");
}

int main() {
  //freopen("input.txt","r",stdin);
  solve();
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

You are given an undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M. In addition, each vertex has a label, `A` or `B`. The label of Vertex i is s_i. Edge i bidirectionally connects vertex a_i and b_i.

The phantom thief Nusook likes to choose some vertex as the startpoint and traverse an edge zero or more times. Today, he will make a string after traveling as above, by placing the labels of the visited vertices in the order visited, beginning from the startpoint.

For example, in a graph where Vertex 1 has the label `A` and Vertex 2 has the label `B`, if Nusook travels along the path 1 \rightarrow 2 \rightarrow 1 \rightarrow 2 \rightarrow 2, the resulting string is `ABABB`.

Determine if Nusook can make all strings consisting of `A` and `B`.

Constraints

* 2 \leq N \leq 2 \times 10^{5}
* 1 \leq M \leq 2 \times 10^{5}
* |s| = N
* s_i is `A` or `B`.
* 1 \leq a_i, b_i \leq N
* The given graph may NOT be simple or connected.

Input

Input is given from Standard Input in the following format:


N M
s
a_1 b_1
:
a_{M} b_{M}


Output

If Nusook can make all strings consisting of `A` and `B`, print `Yes`; otherwise, print `No`.

Examples

Input

2 3
AB
1 1
1 2
2 2


Output

Yes


Input

4 3
ABAB
1 2
2 3
3 1


Output

No


Input

13 23
ABAAAABBBBAAB
7 1
10 6
1 11
2 10
2 8
2 11
11 12
8 3
7 12
11 2
13 13
11 9
4 1
9 7
9 6
8 13
8 6
4 10
8 7
4 3
2 1
8 12
6 9


Output

Yes


Input

13 17
BBABBBAABABBA
7 1
7 9
11 12
3 9
11 9
2 1
11 5
12 11
10 8
1 11
1 8
7 7
9 10
8 8
8 12
6 2
13 11


Output

No

### Response

```cpp
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <assert.h>
#include <memory.h>
#include <queue>
using namespace std;

#define N 200002

vector<int >edge[N];

char s[N];
int vis[N][4];
char t[4] ={'A','A','B','B'};
bool dfs(int node,int ord) {
  if(vis[node][ord]==1) {
    return true;
  }
  vis[node][ord] = 1;
  char tt = t[(ord+1)%4];
  for(int i=0;i<edge[node].size();++i) {
    int ch = edge[node][i];
    char g = s[ch];
    if(tt==g) {
      if(dfs(ch,(ord+1)%4)) {
        return true;
      }
    }
  }
  vis[node][ord] = 2;
  return false;
}
void solve() {
  int n,m;scanf("%d %d",&n,&m);
  scanf(" %s",s+1);
  for(int i=m;i;--i) {
    int a,b;scanf("%d %d",&a,&b);
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  for(int i=1;i<=n;++i) {
    if(!vis[i][0] && s[i]=='A') {
      if(dfs(i,0)) {
        printf("Yes\n" );
        return;
      }
    }
  }
  printf("No\n");
}

int main() {
  //freopen("input.txt","r",stdin);
  solve();
  return 0;
}

```