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

const int N = 200005;

vector<int> adj[N];
int n, m;
string s;
int cnt[N][2]; // 0=>A, 1=>B
bool vis[N];

void bfs() {
	queue<int> Q;
	for (int i = 0; i < n; i++) {
		if (cnt[i][0] == 0 || cnt[i][1] == 0) {
			Q.push(i); // invalid nodes
			vis[i] = 1;
		}
	}
	while (!Q.empty()) {
		int cur = Q.front(); // this node will be removed
		Q.pop();
//		for (int i  = 0; i < adj[cur].size(); i++) {
//			int to = adj[cur][i];
//		}
		for (auto to : adj[cur]) {
			if (vis[to])
				continue;
			int curColor = s[cur] - 'A';
			cnt[to][curColor]--;
			if (cnt[to][0] == 0 || cnt[to][1] == 0) {
				Q.push(to);
				vis[to] = 1;
			}
		}
	}
}

int main() {
	cin >> n >> m;
	cin >> s;
	for (int i = 0; i < m; i++) {
		int f, t;
		cin >> f >> t;
		f--, t--;
		adj[f].push_back(t);
		adj[t].push_back(f);
		cnt[f][s[t] - 'A']++;
		cnt[t][s[f] - 'A']++;
	}
	bfs();
	for (int i = 0; i < n; i++) {
		if (!vis[i]) {
			puts("Yes");
			return 0;
		}
	}
	puts("No");
	return 0;
}

### 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 = 200005;

vector<int> adj[N];
int n, m;
string s;
int cnt[N][2]; // 0=>A, 1=>B
bool vis[N];

void bfs() {
	queue<int> Q;
	for (int i = 0; i < n; i++) {
		if (cnt[i][0] == 0 || cnt[i][1] == 0) {
			Q.push(i); // invalid nodes
			vis[i] = 1;
		}
	}
	while (!Q.empty()) {
		int cur = Q.front(); // this node will be removed
		Q.pop();
//		for (int i  = 0; i < adj[cur].size(); i++) {
//			int to = adj[cur][i];
//		}
		for (auto to : adj[cur]) {
			if (vis[to])
				continue;
			int curColor = s[cur] - 'A';
			cnt[to][curColor]--;
			if (cnt[to][0] == 0 || cnt[to][1] == 0) {
				Q.push(to);
				vis[to] = 1;
			}
		}
	}
}

int main() {
	cin >> n >> m;
	cin >> s;
	for (int i = 0; i < m; i++) {
		int f, t;
		cin >> f >> t;
		f--, t--;
		adj[f].push_back(t);
		adj[t].push_back(f);
		cnt[f][s[t] - 'A']++;
		cnt[t][s[f] - 'A']++;
	}
	bfs();
	for (int i = 0; i < n; i++) {
		if (!vis[i]) {
			puts("Yes");
			return 0;
		}
	}
	puts("No");
	return 0;
}

```