#include <bits/stdc++.h>

using namespace std;

const int MX = 200000;

vector<int> G[MX];
int neigh[MX][2];
char s[MX + 1];
bool deleted[MX];

void dfs(int v) {
	if (deleted[v]) return;
	deleted[v] = true;
	for (int u : G[v]) {
		neigh[u][s[v] - 'A']--;
		if (neigh[u][s[v] - 'A'] == 0) dfs(u);
	}
}

int main() {
	int n, m;
	ignore = scanf("%d %d %s", &n, &m, s);
	for (int i = 0, u, v; i < m; i++) {
		ignore = scanf("%d %d", &u, &v);
		u--; v--;
		G[u].emplace_back(v);
		G[v].emplace_back(u);
		neigh[u][s[v] - 'A']++;
		neigh[v][s[u] - 'A']++;
	}
	
	for (int i = 0; i < n; i++)
		if (neigh[i][0] == 0 || neigh[i][1] == 0)
			dfs(i);
	
	int v = 0;
	while (v < n && deleted[v]) v++;
	
	printf("%s\n", v == n ? "No" : "Yes");
	
	return 0;
}

### Prompt

Develop a solution in cpp 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;

const int MX = 200000;

vector<int> G[MX];
int neigh[MX][2];
char s[MX + 1];
bool deleted[MX];

void dfs(int v) {
	if (deleted[v]) return;
	deleted[v] = true;
	for (int u : G[v]) {
		neigh[u][s[v] - 'A']--;
		if (neigh[u][s[v] - 'A'] == 0) dfs(u);
	}
}

int main() {
	int n, m;
	ignore = scanf("%d %d %s", &n, &m, s);
	for (int i = 0, u, v; i < m; i++) {
		ignore = scanf("%d %d", &u, &v);
		u--; v--;
		G[u].emplace_back(v);
		G[v].emplace_back(u);
		neigh[u][s[v] - 'A']++;
		neigh[v][s[u] - 'A']++;
	}
	
	for (int i = 0; i < n; i++)
		if (neigh[i][0] == 0 || neigh[i][1] == 0)
			dfs(i);
	
	int v = 0;
	while (v < n && deleted[v]) v++;
	
	printf("%s\n", v == n ? "No" : "Yes");
	
	return 0;
}

```