#include <cstdio>
#include <queue>
const int N = 200005;
int n, m, cnt[2][N], head[N], tot, vis[N];
char buf[N];
std::queue<int> q;
struct edge { int to, nxt;} e[ N << 1];
void addedge(int x, int y) {
	e[++tot] = (edge) { y, head[x] }; head[x] = tot;
}
int main() {
	std::scanf("%d%d", &n, &m);
	std::scanf("%s", buf + 1);
	for (int i = 1; i <= m; i++) {
		int x, y;
		std::scanf("%d%d", &x, &y);
		if (x == y) cnt[buf[x] - 'A'][x] = 0x3f3f3f3f;
		else addedge(x, y), addedge(y, x), ++cnt[buf[x] - 'A'][y], ++cnt[buf[y] - 'A'][x];
	}
	for (int i = 1; i <= n; i++) if (!(cnt[0][i] && cnt[1][i])) vis[i] = 1, q.push(i);
	while (!q.empty()) {
		int tmp = q.front(); q.pop();
		for (int i = head[tmp]; i; i = e[i].nxt) if (!vis[e[i].to]) {
			--cnt[buf[tmp] - 'A'][e[i].to];
			if (!(cnt[0][e[i].to] && cnt[1][e[i].to])) vis[e[i].to] = 1, q.push(e[i].to);
		}
	}
	int ans = 1;
	for (int i = 1; i <= n; i++) ans &= vis[i];
	std::puts(ans ? "No" : "Yes");
	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 <cstdio>
#include <queue>
const int N = 200005;
int n, m, cnt[2][N], head[N], tot, vis[N];
char buf[N];
std::queue<int> q;
struct edge { int to, nxt;} e[ N << 1];
void addedge(int x, int y) {
	e[++tot] = (edge) { y, head[x] }; head[x] = tot;
}
int main() {
	std::scanf("%d%d", &n, &m);
	std::scanf("%s", buf + 1);
	for (int i = 1; i <= m; i++) {
		int x, y;
		std::scanf("%d%d", &x, &y);
		if (x == y) cnt[buf[x] - 'A'][x] = 0x3f3f3f3f;
		else addedge(x, y), addedge(y, x), ++cnt[buf[x] - 'A'][y], ++cnt[buf[y] - 'A'][x];
	}
	for (int i = 1; i <= n; i++) if (!(cnt[0][i] && cnt[1][i])) vis[i] = 1, q.push(i);
	while (!q.empty()) {
		int tmp = q.front(); q.pop();
		for (int i = head[tmp]; i; i = e[i].nxt) if (!vis[e[i].to]) {
			--cnt[buf[tmp] - 'A'][e[i].to];
			if (!(cnt[0][e[i].to] && cnt[1][e[i].to])) vis[e[i].to] = 1, q.push(e[i].to);
		}
	}
	int ans = 1;
	for (int i = 1; i <= n; i++) ans &= vis[i];
	std::puts(ans ? "No" : "Yes");
	return 0;
} 
```