#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int const N = 200000;
vector<int> g[N + 1];
char cl[N + 2], z[] = "AABB";
int vis[N + 1][4];

bool go(int v, int j){
	vis[v][j] = 1;
	int nj = j + 1;
	if (nj == 4)nj = 0;
	f(i, 0, g[v].size()){
		int u = g[v][i];
		if (cl[u] != z[nj])continue;
		if (vis[u][nj] == 1)return true;
		if (!vis[u][nj] && go(u, nj))return true;
	}
	vis[v][j] = 2;
	return false;
}

int main(){
	int n, m;
	scanf("%d%d%s", &n, &m, cl + 1);
	f(i, 0, m){
		int a, b;
		scanf("%d%d", &a, &b);
		g[a].push_back(b), g[b].push_back(a);
	}
	f(i, 1, n + 1)if (cl[i] == 'B' && !vis[i][2] && !vis[i][3] && go(i, 3)) { printf("Yes\n"); return 0; }
	printf("No\n");
}
### Prompt

Please formulate a CPP solution to 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 <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int const N = 200000;
vector<int> g[N + 1];
char cl[N + 2], z[] = "AABB";
int vis[N + 1][4];

bool go(int v, int j){
	vis[v][j] = 1;
	int nj = j + 1;
	if (nj == 4)nj = 0;
	f(i, 0, g[v].size()){
		int u = g[v][i];
		if (cl[u] != z[nj])continue;
		if (vis[u][nj] == 1)return true;
		if (!vis[u][nj] && go(u, nj))return true;
	}
	vis[v][j] = 2;
	return false;
}

int main(){
	int n, m;
	scanf("%d%d%s", &n, &m, cl + 1);
	f(i, 0, m){
		int a, b;
		scanf("%d%d", &a, &b);
		g[a].push_back(b), g[b].push_back(a);
	}
	f(i, 1, n + 1)if (cl[i] == 'B' && !vis[i][2] && !vis[i][3] && go(i, 3)) { printf("Yes\n"); return 0; }
	printf("No\n");
}
```