#include<bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=200005;
int n,m;
char s[maxn];
vector<int>edge[maxn];
queue<int>q;
int cntA[maxn],cntB[maxn];
bool bad[maxn];

int main(){
	scanf("%d%d%s",&n,&m,s+1);
	REP(i,m){
		int u,v;
		scanf("%d%d",&u,&v);
		edge[u].pb(v);
		edge[v].pb(u);
		if(s[u]=='A')cntA[v]++;
		else cntB[v]++;
		if(s[v]=='A')cntA[u]++;
		else cntB[u]++; 
	}
	REP(i,n)if(!cntA[i]||!cntB[i])q.push(i),bad[i]=true;
	while(!q.empty()){
		int x=q.front();q.pop();
		rep(i,int(edge[x].size())){
			int y=edge[x][i];
			if(s[x]=='A')cntA[y]--;
			else cntB[y]--;
			if(!cntA[y]||!cntB[y]){
				if(!bad[y]){
					bad[y]=true;
					q.push(y);
				}
			}
		}
	}
	REP(i,n)if(!bad[i]){
		puts("Yes");
		return 0;
	}
	puts("No");
	return 0;
} 
### Prompt

Create a solution in cpp for 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;

#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=200005;
int n,m;
char s[maxn];
vector<int>edge[maxn];
queue<int>q;
int cntA[maxn],cntB[maxn];
bool bad[maxn];

int main(){
	scanf("%d%d%s",&n,&m,s+1);
	REP(i,m){
		int u,v;
		scanf("%d%d",&u,&v);
		edge[u].pb(v);
		edge[v].pb(u);
		if(s[u]=='A')cntA[v]++;
		else cntB[v]++;
		if(s[v]=='A')cntA[u]++;
		else cntB[u]++; 
	}
	REP(i,n)if(!cntA[i]||!cntB[i])q.push(i),bad[i]=true;
	while(!q.empty()){
		int x=q.front();q.pop();
		rep(i,int(edge[x].size())){
			int y=edge[x][i];
			if(s[x]=='A')cntA[y]--;
			else cntB[y]--;
			if(!cntA[y]||!cntB[y]){
				if(!bad[y]){
					bad[y]=true;
					q.push(y);
				}
			}
		}
	}
	REP(i,n)if(!bad[i]){
		puts("Yes");
		return 0;
	}
	puts("No");
	return 0;
} 
```