#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 2000005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

inline int read()
{
	int x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}

int n,m,x[maxn][2],inq[maxn],cnt;
char s[maxn];
vector <int> mp[maxn];
queue <int> q;

int main()
{
	n=read(); m=read(); scanf("%s",s+1);
	rep(i,1,m)
	{
		int u=read(),v=read();
		mp[u].pb(v); mp[v].pb(u); x[u][s[v]-'A']++; x[v][s[u]-'A']++;
	}
	rep(i,1,n) if(x[i][0]==0||x[i][1]==0) q.push(i),inq[i]=1;
	while(!q.empty())
	{
		int u=q.front(); q.pop();
		for(int i=0;i<mp[u].size();i++)
		{
			int v=mp[u][i]; if(inq[v]) continue;
			x[v][s[u]-'A']--;
			if(x[v][0]==0||x[v][1]==0) q.push(v),inq[v]=1;
		}
	}
	rep(i,1,n) cnt+=inq[i];
	if(cnt!=n) puts("Yes"); else puts("No");
	return 0;
}
### Prompt

Your challenge is to write 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
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 2000005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

inline int read()
{
	int x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}

int n,m,x[maxn][2],inq[maxn],cnt;
char s[maxn];
vector <int> mp[maxn];
queue <int> q;

int main()
{
	n=read(); m=read(); scanf("%s",s+1);
	rep(i,1,m)
	{
		int u=read(),v=read();
		mp[u].pb(v); mp[v].pb(u); x[u][s[v]-'A']++; x[v][s[u]-'A']++;
	}
	rep(i,1,n) if(x[i][0]==0||x[i][1]==0) q.push(i),inq[i]=1;
	while(!q.empty())
	{
		int u=q.front(); q.pop();
		for(int i=0;i<mp[u].size();i++)
		{
			int v=mp[u][i]; if(inq[v]) continue;
			x[v][s[u]-'A']--;
			if(x[v][0]==0||x[v][1]==0) q.push(v),inq[v]=1;
		}
	}
	rep(i,1,n) cnt+=inq[i];
	if(cnt!=n) puts("Yes"); else puts("No");
	return 0;
}
```