#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
const int N = 2e5 + 5;
int n, m, cnt[N][2];
char str[N];
bool tag[N], del[N];
vector<int> G[N];
queue<int> Q;
int main() {
	scanf("%d%d%s", &n, &m, str + 1);
	for (int i = 1; i <= n; ++i)
		tag[i] = (str[i] == 'A') ? 1 : 0;
	for (int i = 1; i <= m; ++i) {
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for (int i = 1; i <= n; ++i)
		for (int to : G[i])
			++cnt[i][tag[to]];
	while (Q.size())
		Q.pop();
	for (int i = 1; i <= n; ++i)
		if (!cnt[i][0] || !cnt[i][1])
			Q.push(i), del[i] = 1;
	while (Q.size()) {
		int u = Q.front();
		Q.pop();
		for (int to : G[u]) {
			if (del[to])
				continue ;
			--cnt[to][tag[u]];
			if (!cnt[to][0] || !cnt[to][1]) {
				Q.push(to);
				del[to] = 1;
			}
		}
	}
	bool ans = 0;
	for (int i = 1; i <= n; ++i)
		ans |= !del[i];
	if (ans)
		printf("Yes\n");
	else
		printf("No\n");
	return 0;
}
### Prompt

Your task is to create 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 <cstdio>
#include <vector>
#include <queue>
using namespace std;
const int N = 2e5 + 5;
int n, m, cnt[N][2];
char str[N];
bool tag[N], del[N];
vector<int> G[N];
queue<int> Q;
int main() {
	scanf("%d%d%s", &n, &m, str + 1);
	for (int i = 1; i <= n; ++i)
		tag[i] = (str[i] == 'A') ? 1 : 0;
	for (int i = 1; i <= m; ++i) {
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for (int i = 1; i <= n; ++i)
		for (int to : G[i])
			++cnt[i][tag[to]];
	while (Q.size())
		Q.pop();
	for (int i = 1; i <= n; ++i)
		if (!cnt[i][0] || !cnt[i][1])
			Q.push(i), del[i] = 1;
	while (Q.size()) {
		int u = Q.front();
		Q.pop();
		for (int to : G[u]) {
			if (del[to])
				continue ;
			--cnt[to][tag[u]];
			if (!cnt[to][0] || !cnt[to][1]) {
				Q.push(to);
				del[to] = 1;
			}
		}
	}
	bool ans = 0;
	for (int i = 1; i <= n; ++i)
		ans |= !del[i];
	if (ans)
		printf("Yes\n");
	else
		printf("No\n");
	return 0;
}
```