#include<cstdio>
#include<vector>
#include<algorithm>
#define pb push_back
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
using namespace std;

const int N=200010;
char s[N];
int n,m,x,y,d[N][2],vis[N];
vector<int>v[N],q;

int main(){
	scanf("%d%d%s",&n,&m,s+1);
	for (int i=m; i; i--){
		scanf("%d%d",&x,&y); v[x].pb(y); v[y].pb(x);
		++d[x][s[y]-'A']; ++d[y][s[x]-'A'];
	}
	rep(i,1,n) if (!d[i][0] || !d[i][1]) q.pb(i),vis[i]=1;
	for (int j=0; j<(int)q.size(); j++){
		int sz=v[q[j]].size()-1;
		rep(i,0,sz){
			int p=v[q[j]][i];
			if (!--d[p][s[q[j]]-'A'] && !vis[p]) q.pb(p),vis[p]=1;
		}
	}
	if ((int)q.size()<n) puts("Yes"); else puts("No");
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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<cstdio>
#include<vector>
#include<algorithm>
#define pb push_back
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
using namespace std;

const int N=200010;
char s[N];
int n,m,x,y,d[N][2],vis[N];
vector<int>v[N],q;

int main(){
	scanf("%d%d%s",&n,&m,s+1);
	for (int i=m; i; i--){
		scanf("%d%d",&x,&y); v[x].pb(y); v[y].pb(x);
		++d[x][s[y]-'A']; ++d[y][s[x]-'A'];
	}
	rep(i,1,n) if (!d[i][0] || !d[i][1]) q.pb(i),vis[i]=1;
	for (int j=0; j<(int)q.size(); j++){
		int sz=v[q[j]].size()-1;
		rep(i,0,sz){
			int p=v[q[j]][i];
			if (!--d[p][s[q[j]]-'A'] && !vis[p]) q.pb(p),vis[p]=1;
		}
	}
	if ((int)q.size()<n) puts("Yes"); else puts("No");
	return 0;
}
```