#include <bits/stdc++.h>
using namespace std;
typedef pair <int, int> pii;

const int N = 2e5 + 10;
int n, m, mark[N], deg[2][N];
vector <int> vec, adj[N];
string s;

void dfs(int v) {
	mark[v] = 0;
	for (int i : adj[v])
		if (mark[i]) {
			deg[s[v] - 'A'][i]--;
			if (!deg[s[v] - 'A'][i])
				vec.push_back(i);
		}
}

int main() {
	fill(mark, mark + N, 1);
	
	cin >> n >> m >> s;
	for (int i = 0; i < m; i++) {
		int v, u;
		cin >> v >> u;
		v--;
		u--;
		if (v == u) {
			deg[s[v] - 'A'][v]++;
			continue;
		}
		adj[v].push_back(u);
		adj[u].push_back(v);
		deg[s[u] - 'A'][v]++, deg[s[v] - 'A'][u]++;
	}
	
	for (int i = 0; i < n; i++)
		if (!deg[0][i] || !deg[1][i])
			vec.push_back(i);
	
	while (vec.size()) {
		int tmp = vec.back();
		vec.pop_back();
		dfs(tmp);
	}
	
	for (int i = 0; i < n; i++)
		if (mark[i])
//			cout << i + 1 << " ";
			return cout << "Yes", 0;
	cout << "No";
}

### 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;
typedef pair <int, int> pii;

const int N = 2e5 + 10;
int n, m, mark[N], deg[2][N];
vector <int> vec, adj[N];
string s;

void dfs(int v) {
	mark[v] = 0;
	for (int i : adj[v])
		if (mark[i]) {
			deg[s[v] - 'A'][i]--;
			if (!deg[s[v] - 'A'][i])
				vec.push_back(i);
		}
}

int main() {
	fill(mark, mark + N, 1);
	
	cin >> n >> m >> s;
	for (int i = 0; i < m; i++) {
		int v, u;
		cin >> v >> u;
		v--;
		u--;
		if (v == u) {
			deg[s[v] - 'A'][v]++;
			continue;
		}
		adj[v].push_back(u);
		adj[u].push_back(v);
		deg[s[u] - 'A'][v]++, deg[s[v] - 'A'][u]++;
	}
	
	for (int i = 0; i < n; i++)
		if (!deg[0][i] || !deg[1][i])
			vec.push_back(i);
	
	while (vec.size()) {
		int tmp = vec.back();
		vec.pop_back();
		dfs(tmp);
	}
	
	for (int i = 0; i < n; i++)
		if (mark[i])
//			cout << i + 1 << " ";
			return cout << "Yes", 0;
	cout << "No";
}

```