#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;

using namespace std;
#define N 200010
vector<int> v[N];
int n,m;string s;
int cnt[N][2];
bool vis[N];
int main(){
    cin>>n>>m>>s;
    lol(i,m){
	int a,b;cin>>a>>b;a--,b--;
	v[a].push_back(b);
	v[b].push_back(a);
    }
    queue<int> Q;
    lol(i,n){
	vis[i]=0;
	lol(j,2)cnt[i][j]=0;
	for(auto x:v[i]){
	    if(s[x]=='A')cnt[i][0]++;
	    if(s[x]=='B')cnt[i][1]++;
	}
	if(cnt[i][0]==0||cnt[i][1]==0){
	    Q.push(i);
	    vis[i]=1;
	}
    }
    while(!Q.empty()){
	int x=Q.front();Q.pop();
	for(auto y:v[x]){
	    if(vis[y]==1)continue;
	    if(s[x]=='A')cnt[y][0]--;
	    if(s[x]=='B')cnt[y][1]--;
	    if(cnt[y][0]<=0||cnt[y][1]<=0){
		Q.push(y);
		vis[y]=1;
	    }
	}
    }
    bool ans=1;
    lol(i,n)ans&=vis[i];
    cout<<(ans?"No":"Yes")<<endl;
    return 0;
}

### Prompt

Please formulate 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
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;

using namespace std;
#define N 200010
vector<int> v[N];
int n,m;string s;
int cnt[N][2];
bool vis[N];
int main(){
    cin>>n>>m>>s;
    lol(i,m){
	int a,b;cin>>a>>b;a--,b--;
	v[a].push_back(b);
	v[b].push_back(a);
    }
    queue<int> Q;
    lol(i,n){
	vis[i]=0;
	lol(j,2)cnt[i][j]=0;
	for(auto x:v[i]){
	    if(s[x]=='A')cnt[i][0]++;
	    if(s[x]=='B')cnt[i][1]++;
	}
	if(cnt[i][0]==0||cnt[i][1]==0){
	    Q.push(i);
	    vis[i]=1;
	}
    }
    while(!Q.empty()){
	int x=Q.front();Q.pop();
	for(auto y:v[x]){
	    if(vis[y]==1)continue;
	    if(s[x]=='A')cnt[y][0]--;
	    if(s[x]=='B')cnt[y][1]--;
	    if(cnt[y][0]<=0||cnt[y][1]<=0){
		Q.push(y);
		vis[y]=1;
	    }
	}
    }
    bool ans=1;
    lol(i,n)ans&=vis[i];
    cout<<(ans?"No":"Yes")<<endl;
    return 0;
}

```