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

int main() {
	int n, m;
	scanf("%d %d", &n, &m);
	string s(n, ' ');
	scanf("%s", s.data());

	vector<vector<int>> graph(n);
	vector<array<int, 2>> cnt(n);

	while (m--) {
		int a, b;
		scanf("%d %d", &a, &b); a--; b--;
		graph[a].push_back(b);
		graph[b].push_back(a);
		cnt[a][s[b]-'A']++;
		cnt[b][s[a]-'A']++;
	}

	vector<int> rem;
	vector<bool> removed(n, false);
	for (int i = 0; i < n; i++)
		if (cnt[i][0] == 0 || cnt[i][1] == 0) {
			removed[i] = true;
			rem.push_back(i);
		}

	for (int i = 0; i < rem.size(); i++) {
		int x = rem[i];

		int inx = s[x] - 'A';
		
		for (int y : graph[x]) {
			if ((--cnt[y][inx]) == 0 && !removed[y]) {
				removed[y] = true;
				rem.push_back(y);
			}
		}
	}

	if (rem.size() == n) printf("No\n");
	else printf("Yes\n");

	return 0;
}
### Prompt

Create a solution in Cpp for 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;

int main() {
	int n, m;
	scanf("%d %d", &n, &m);
	string s(n, ' ');
	scanf("%s", s.data());

	vector<vector<int>> graph(n);
	vector<array<int, 2>> cnt(n);

	while (m--) {
		int a, b;
		scanf("%d %d", &a, &b); a--; b--;
		graph[a].push_back(b);
		graph[b].push_back(a);
		cnt[a][s[b]-'A']++;
		cnt[b][s[a]-'A']++;
	}

	vector<int> rem;
	vector<bool> removed(n, false);
	for (int i = 0; i < n; i++)
		if (cnt[i][0] == 0 || cnt[i][1] == 0) {
			removed[i] = true;
			rem.push_back(i);
		}

	for (int i = 0; i < rem.size(); i++) {
		int x = rem[i];

		int inx = s[x] - 'A';
		
		for (int y : graph[x]) {
			if ((--cnt[y][inx]) == 0 && !removed[y]) {
				removed[y] = true;
				rem.push_back(y);
			}
		}
	}

	if (rem.size() == n) printf("No\n");
	else printf("Yes\n");

	return 0;
}
```