#include <bits/stdc++.h>

using namespace std;

typedef double D;
typedef long long int LL;

#define st first
#define nd second
#define pb push_back
#define PLL pair <LL, LL>
#define PII pair <int, int>

const int N = 2e5 + 7;
const int MX = 1e9 + 7;
const LL INF = 1e18 + 9LL;

int n, m;
char type[N];
int ok[N][2];

bool bad[N];
vector <int> G[N];

int main(){
	scanf("%d %d", &n, &m);
	scanf("%s", type + 1);

	for(int i = 1; i <= m; ++i){
		int u, v;
		scanf("%d %d", &u, &v);
		
		ok[u][type[v] - 'A']++;
		ok[v][type[u] - 'A']++;

		G[u].push_back(v);
		G[v].push_back(u);
	}
	
	queue <int> Q;
	for(int i = 1; i <= n; ++i)
		if(ok[i][0] == 0 || ok[i][1] == 0){
			bad[i] = true;
			Q.push(i);
		}
	
	while(!Q.empty()){
		int u = Q.front();
		Q.pop();
		
		for(int v: G[u]){
			if(bad[v])
				continue;
			
			ok[v][type[u] - 'A']--;
			if(ok[v][0] == 0 || ok[v][1] == 0){
				bad[v] = true;
				Q.push(v);
			}
		}
	}
	
	for(int i = 1; i <= n; ++i)
		if(!bad[i]){
			puts("Yes");
			return 0;
		}
	
	puts("No");
	return 0;
}
### Prompt

Your task is to create a Cpp solution 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;

typedef double D;
typedef long long int LL;

#define st first
#define nd second
#define pb push_back
#define PLL pair <LL, LL>
#define PII pair <int, int>

const int N = 2e5 + 7;
const int MX = 1e9 + 7;
const LL INF = 1e18 + 9LL;

int n, m;
char type[N];
int ok[N][2];

bool bad[N];
vector <int> G[N];

int main(){
	scanf("%d %d", &n, &m);
	scanf("%s", type + 1);

	for(int i = 1; i <= m; ++i){
		int u, v;
		scanf("%d %d", &u, &v);
		
		ok[u][type[v] - 'A']++;
		ok[v][type[u] - 'A']++;

		G[u].push_back(v);
		G[v].push_back(u);
	}
	
	queue <int> Q;
	for(int i = 1; i <= n; ++i)
		if(ok[i][0] == 0 || ok[i][1] == 0){
			bad[i] = true;
			Q.push(i);
		}
	
	while(!Q.empty()){
		int u = Q.front();
		Q.pop();
		
		for(int v: G[u]){
			if(bad[v])
				continue;
			
			ok[v][type[u] - 'A']--;
			if(ok[v][0] == 0 || ok[v][1] == 0){
				bad[v] = true;
				Q.push(v);
			}
		}
	}
	
	for(int i = 1; i <= n; ++i)
		if(!bad[i]){
			puts("Yes");
			return 0;
		}
	
	puts("No");
	return 0;
}
```