#include <bits/stdc++.h>
using namespace std;
const int MX = 200005;

char S[MX];

int na[MX];
int nb[MX];

vector<int> neigh[MX];
bool destroyed[MX];

void destroy(int v) {
	destroyed[v] = true;
	for (int u : neigh[v]) {
		if (destroyed[u]) continue;
		if (S[v] == 'A') {
			na[u]--;
		}
		else {
			nb[u]--;
		}
		
		if (na[u] == 0 || nb[u] == 0) {
			destroy(u);
		}
	}
}

int main()
{
	int n, m;
	ignore = scanf("%d %d", &n, &m);
	ignore = scanf(" %s", S + 1);
	
	for (int i = 0; i < m; i++) {
		int u, v;
		ignore = scanf("%d %d", &u, &v);
		neigh[u].push_back(v);
		neigh[v].push_back(u);
		
		if (S[v] == 'A') {
			na[u]++;
		}
		else {
			nb[u]++;
		}
		
		if (u != v) {		
			if (S[u] == 'A') {
				na[v]++;
			}
			else {
				nb[v]++;
			}
		}
	}

	for (int v = 1; v <= n; v++) {
		if (destroyed[v]) continue;
		if (na[v] == 0 || nb[v] == 0) {
			destroy(v);
		}
	}
	
	for (int v = 1; v <= n; v++) {
		if (!destroyed[v]) {
			printf("Yes\n");
			return 0;
		}
	}
	
	printf("No\n");

	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 MX = 200005;

char S[MX];

int na[MX];
int nb[MX];

vector<int> neigh[MX];
bool destroyed[MX];

void destroy(int v) {
	destroyed[v] = true;
	for (int u : neigh[v]) {
		if (destroyed[u]) continue;
		if (S[v] == 'A') {
			na[u]--;
		}
		else {
			nb[u]--;
		}
		
		if (na[u] == 0 || nb[u] == 0) {
			destroy(u);
		}
	}
}

int main()
{
	int n, m;
	ignore = scanf("%d %d", &n, &m);
	ignore = scanf(" %s", S + 1);
	
	for (int i = 0; i < m; i++) {
		int u, v;
		ignore = scanf("%d %d", &u, &v);
		neigh[u].push_back(v);
		neigh[v].push_back(u);
		
		if (S[v] == 'A') {
			na[u]++;
		}
		else {
			nb[u]++;
		}
		
		if (u != v) {		
			if (S[u] == 'A') {
				na[v]++;
			}
			else {
				nb[v]++;
			}
		}
	}

	for (int v = 1; v <= n; v++) {
		if (destroyed[v]) continue;
		if (na[v] == 0 || nb[v] == 0) {
			destroy(v);
		}
	}
	
	for (int v = 1; v <= n; v++) {
		if (!destroyed[v]) {
			printf("Yes\n");
			return 0;
		}
	}
	
	printf("No\n");

	return 0;
}

```