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
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#define SF scanf
#define PF printf
#define MAXN 200010
using namespace std;
typedef long long ll;
int n,m;
int cnt[MAXN][2];
char s[MAXN];
bool del[MAXN];
vector<int> a[MAXN];
void dfs(int x){
	del[x]=1;
	for(int i=0;i<a[x].size();i++){
		int u=a[x][i];
		if(del[u]==0){
			cnt[u][s[x]-'A']--;
			if(cnt[u][s[x]-'A']==0)
				dfs(u);
		}
	}
}
int main(){
	SF("%d%d",&n,&m);
	SF("%s",s+1);
	int u,v;
	for(int i=1;i<=m;i++){
		SF("%d%d",&u,&v);
		a[u].push_back(v);
		a[v].push_back(u);
		cnt[v][s[u]-'A']++;
		cnt[u][s[v]-'A']++;	
	}
	for(int i=1;i<=n;i++)
		if(del[i]==0&&(cnt[i][0]==0||cnt[i][1]==0))
			dfs(i);
	for(int i=1;i<=n;i++)
		if(del[i]==0){
			PF("Yes");
			return 0;
		}
	PF("No");
}