#include <bits/stdc++.h>

#define pb push_back
#define For(i, j, k) for (int i = j; i <= k; i++)

using namespace std;

const int N = 2e5 + 10;

int n, m;
int dega[N], degb[N];
bool bad[N];
char S[N];
vector<int> G[N];

int main() {

	scanf("%d%d", &n, &m);
	scanf("%s", S + 1);
	For(i, 1, m) {
		int u, v;
		scanf("%d%d", &u, &v);
		G[v].pb(u), G[u].pb(v);
		if (S[v] == 'A') ++dega[u];
		else ++degb[u];
		if (S[u] == 'A') ++dega[v];
		else ++degb[v];
	}

	queue<int> q;
	For(i, 1, n) if (!dega[i] || !degb[i]) {
		bad[i] = true;
		q.push(i);
	}

	while (!q.empty()) {
		int o = q.front(); q.pop();
		for (int v : G[o]) if (!bad[v]) {
			if (S[o] == 'A') --dega[v];
			else --degb[v];
			if (!dega[v] || !degb[v]) q.push(v), bad[v] = true;
		}
	}

	For(i, 1, n) if (!bad[i]) {
		puts("Yes");
		return 0;
	}
	puts("No");

	return 0;
}
### Prompt

Generate 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>

#define pb push_back
#define For(i, j, k) for (int i = j; i <= k; i++)

using namespace std;

const int N = 2e5 + 10;

int n, m;
int dega[N], degb[N];
bool bad[N];
char S[N];
vector<int> G[N];

int main() {

	scanf("%d%d", &n, &m);
	scanf("%s", S + 1);
	For(i, 1, m) {
		int u, v;
		scanf("%d%d", &u, &v);
		G[v].pb(u), G[u].pb(v);
		if (S[v] == 'A') ++dega[u];
		else ++degb[u];
		if (S[u] == 'A') ++dega[v];
		else ++degb[v];
	}

	queue<int> q;
	For(i, 1, n) if (!dega[i] || !degb[i]) {
		bad[i] = true;
		q.push(i);
	}

	while (!q.empty()) {
		int o = q.front(); q.pop();
		for (int v : G[o]) if (!bad[v]) {
			if (S[o] == 'A') --dega[v];
			else --degb[v];
			if (!dega[v] || !degb[v]) q.push(v), bad[v] = true;
		}
	}

	For(i, 1, n) if (!bad[i]) {
		puts("Yes");
		return 0;
	}
	puts("No");

	return 0;
}
```