#include <bits/stdc++.h>
using namespace std;

const int N = 2e5 + 5;

queue <int> q;

int du[N][2], n, m, x, y;

bool vis[N];

vector <int> G[N];

char s[N];

int main() {
	scanf("%d%d", &n, &m);
	scanf("%s", s + 1);
	for(int i = 1; i <= m; ++ i) {
		scanf("%d%d", &x, &y);
		++ du[x][s[y] - 'A'];
		G[y].push_back(x);
		++ du[y][s[x] - 'A'];
		G[x].push_back(y);
	}
	for(int i = 1; i <= n; ++ i) {
		if(!du[i][0] || !du[i][1]) {
			q.push(i);
			vis[i] = 1;
		}
	}
	while(!q.empty()) {
		int ind = q.front();
		q.pop();
		for(int i = 0; i < (int) G[ind].size(); ++ i) {
			if(-- du[G[ind][i]][s[ind] - 'A'] == 0) {
				if(!
				vis[G[ind][i]]) {
					vis[G[ind][i]] = 1;
					q.push(G[ind][i]);
				}
			}
		}
	}
	for(int i = 1; i <= n; ++ i) if(!vis[i]) return 0 * puts("Yes");
	puts("No");
}
### Prompt

Please create a solution in cpp 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;

const int N = 2e5 + 5;

queue <int> q;

int du[N][2], n, m, x, y;

bool vis[N];

vector <int> G[N];

char s[N];

int main() {
	scanf("%d%d", &n, &m);
	scanf("%s", s + 1);
	for(int i = 1; i <= m; ++ i) {
		scanf("%d%d", &x, &y);
		++ du[x][s[y] - 'A'];
		G[y].push_back(x);
		++ du[y][s[x] - 'A'];
		G[x].push_back(y);
	}
	for(int i = 1; i <= n; ++ i) {
		if(!du[i][0] || !du[i][1]) {
			q.push(i);
			vis[i] = 1;
		}
	}
	while(!q.empty()) {
		int ind = q.front();
		q.pop();
		for(int i = 0; i < (int) G[ind].size(); ++ i) {
			if(-- du[G[ind][i]][s[ind] - 'A'] == 0) {
				if(!
				vis[G[ind][i]]) {
					vis[G[ind][i]] = 1;
					q.push(G[ind][i]);
				}
			}
		}
	}
	for(int i = 1; i <= n; ++ i) if(!vis[i]) return 0 * puts("Yes");
	puts("No");
}
```