#include <bits/stdc++.h>
using namespace std;
#define all(V) V.begin(),V.end()
using ll = long long;
const ll MOD = 1000000007;

vector<int> G[200010];
int ABC[200010][2];
bool er[200010];

int main() {
	for (int i = 0;i < 200010;i++)ABC[i][0] = ABC[i][1] = er[i] = 0;
	int N, M, a, b;
	cin >> N >> M;
	string S;
	cin >> S;
	for (int i = 0;i < M;i++) {
		cin >> a >> b;
		a--;b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}

	queue<int> Q;
	for (int i = 0;i < N;i++) {
		for (int x : G[i]) {
			if (S[x] == 'A') {
				ABC[i][0]++;
			}
			if (S[x] == 'B') {
				ABC[i][1]++;
			}
		}
		if (ABC[i][0] == 0 || ABC[i][1] == 0) {
			Q.push(i);
			er[i] = 1;
		}
	}
	bool ab = 0;
	while (!Q.empty()) {
		a = Q.front();
		Q.pop();
		if (S[a] == 'A')ab = 0;
		else ab = 1;
		for (int x : G[a]) {
			ABC[x][ab]--;
			if (ABC[x][ab] == 0 && !er[x]) {
				Q.push(x);
				er[x] = 1;
			}
		}
	}
	for (int i = 0;i < N;i++) {
		if (!er[i]) {
			cout << "Yes" << endl;
			return 0;
		}
	}
	cout << "No" << endl;
}
### Prompt

Please provide a Cpp coded solution 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 <bits/stdc++.h>
using namespace std;
#define all(V) V.begin(),V.end()
using ll = long long;
const ll MOD = 1000000007;

vector<int> G[200010];
int ABC[200010][2];
bool er[200010];

int main() {
	for (int i = 0;i < 200010;i++)ABC[i][0] = ABC[i][1] = er[i] = 0;
	int N, M, a, b;
	cin >> N >> M;
	string S;
	cin >> S;
	for (int i = 0;i < M;i++) {
		cin >> a >> b;
		a--;b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}

	queue<int> Q;
	for (int i = 0;i < N;i++) {
		for (int x : G[i]) {
			if (S[x] == 'A') {
				ABC[i][0]++;
			}
			if (S[x] == 'B') {
				ABC[i][1]++;
			}
		}
		if (ABC[i][0] == 0 || ABC[i][1] == 0) {
			Q.push(i);
			er[i] = 1;
		}
	}
	bool ab = 0;
	while (!Q.empty()) {
		a = Q.front();
		Q.pop();
		if (S[a] == 'A')ab = 0;
		else ab = 1;
		for (int x : G[a]) {
			ABC[x][ab]--;
			if (ABC[x][ab] == 0 && !er[x]) {
				Q.push(x);
				er[x] = 1;
			}
		}
	}
	for (int i = 0;i < N;i++) {
		if (!er[i]) {
			cout << "Yes" << endl;
			return 0;
		}
	}
	cout << "No" << endl;
}
```