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

vector<int> G[200200],V[200200];
int N,M; char S[200200];
int C[2][200200],U[200200],K[200200];

int main()
{
	scanf ("%d %d",&N,&M);
	scanf ("%s",S+1);

	for (int i=0;i<M;i++){
		int x,y; scanf ("%d %d",&x,&y);
		C[S[x]-'A'][y]++;
		C[S[y]-'A'][x]++;
		G[x].push_back(y);
		G[y].push_back(x);
	}

	queue<int> Q;
	for (int i=1;i<=N;i++) if (C[0][i] == 0 || C[1][i] == 0){
		Q.push(i); K[i] = 1;
	}
	while (!Q.empty()){
		int x = Q.front(); Q.pop();
		for (auto &y : G[x]) if (C[0][y] && C[1][y] && K[y] == 0){
			C[S[x]-'A'][y]--;
			C[S[y]-'A'][x]--;
			if (C[0][y] == 0 || C[1][y] == 0){
				Q.push(y); K[y] = 1;
			}
		}
	}

	int ans = 0;
	for (int i=1;i<=N;i++) if (K[i]) ans++;
	puts(ans<N?"Yes":"No");
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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

vector<int> G[200200],V[200200];
int N,M; char S[200200];
int C[2][200200],U[200200],K[200200];

int main()
{
	scanf ("%d %d",&N,&M);
	scanf ("%s",S+1);

	for (int i=0;i<M;i++){
		int x,y; scanf ("%d %d",&x,&y);
		C[S[x]-'A'][y]++;
		C[S[y]-'A'][x]++;
		G[x].push_back(y);
		G[y].push_back(x);
	}

	queue<int> Q;
	for (int i=1;i<=N;i++) if (C[0][i] == 0 || C[1][i] == 0){
		Q.push(i); K[i] = 1;
	}
	while (!Q.empty()){
		int x = Q.front(); Q.pop();
		for (auto &y : G[x]) if (C[0][y] && C[1][y] && K[y] == 0){
			C[S[x]-'A'][y]--;
			C[S[y]-'A'][x]--;
			if (C[0][y] == 0 || C[1][y] == 0){
				Q.push(y); K[y] = 1;
			}
		}
	}

	int ans = 0;
	for (int i=1;i<=N;i++) if (K[i]) ans++;
	puts(ans<N?"Yes":"No");
	return 0;
}
```