#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
inline LL rd() {
	char ch=getchar(); LL i=0,f=1;
	while(!isdigit(ch)) {if(ch=='-')f=-1; ch=getchar();}
	while(isdigit(ch)) {i=(i<<1)+(i<<3)+ch-'0'; ch=getchar();}
	return i*f;
}

const int N=2e5+50, mod=1e9+7;
int n,m,vis[N];
int cnt[N][2],type[N];
vector <int> edge[N];
char ch[N];

int main() {
	n=rd(), m=rd();
	scanf("%s",ch+1);
	for(int i=1;i<=n;i++) type[i]=(ch[i]=='A') ? 0 : 1;
	for(int i=1;i<=m;i++) {
		int x=rd(), y=rd();
		if(x!=y) {
			edge[x].push_back(y);
			edge[y].push_back(x);
		}
		++cnt[x][type[y]];
		++cnt[y][type[x]];
	}
	queue <int> q; int cc=0;
	for(int i=1;i<=n;++i) if(!cnt[i][0] || !cnt[i][1]) q.push(i), vis[i]=1;
	while(!q.empty()) {
		int u=q.front(); q.pop();  ++cc;
		for(auto e:edge[u]) if(!vis[e]) {
			cnt[e][type[u]]--;
			if(!cnt[e][0] || !cnt[e][1]) q.push(e), vis[e]=1;
		}
	}
	if(cc==n) puts("No");
	else puts("Yes");
}
### 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;
typedef long long LL;
inline LL rd() {
	char ch=getchar(); LL i=0,f=1;
	while(!isdigit(ch)) {if(ch=='-')f=-1; ch=getchar();}
	while(isdigit(ch)) {i=(i<<1)+(i<<3)+ch-'0'; ch=getchar();}
	return i*f;
}

const int N=2e5+50, mod=1e9+7;
int n,m,vis[N];
int cnt[N][2],type[N];
vector <int> edge[N];
char ch[N];

int main() {
	n=rd(), m=rd();
	scanf("%s",ch+1);
	for(int i=1;i<=n;i++) type[i]=(ch[i]=='A') ? 0 : 1;
	for(int i=1;i<=m;i++) {
		int x=rd(), y=rd();
		if(x!=y) {
			edge[x].push_back(y);
			edge[y].push_back(x);
		}
		++cnt[x][type[y]];
		++cnt[y][type[x]];
	}
	queue <int> q; int cc=0;
	for(int i=1;i<=n;++i) if(!cnt[i][0] || !cnt[i][1]) q.push(i), vis[i]=1;
	while(!q.empty()) {
		int u=q.front(); q.pop();  ++cc;
		for(auto e:edge[u]) if(!vis[e]) {
			cnt[e][type[u]]--;
			if(!cnt[e][0] || !cnt[e][1]) q.push(e), vis[e]=1;
		}
	}
	if(cc==n) puts("No");
	else puts("Yes");
}
```