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

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

const int N = 2e5 + 25;
vector<int> g[N];
int vis[N][4];
string s;

bool dfs(int v, int i){
	if(vis[v][i] == 2)return false;
	if(vis[v][i] == 1)return true;
	vis[v][i] = 1;
	for(int u : g[v]){
		if(s[u] == 'A'){
			if(i == 0 && dfs(u, 1))return true;
			if(i == 3 && dfs(u, 0))return true;
		}else{
			if(i == 1 && dfs(u, 2))return true;
			if(i == 2 && dfs(u, 3))return true;
		}
	}
	vis[v][i] = 2;
	return false;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout<<setprecision(32);

	int n, m;
	cin>>n>>m>>s;
	for(int i = 0; i < m; i++){
		int u, v;
		cin>>u>>v;
		u--; v--;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	memset(vis, 0, sizeof(vis));
	for(int i = 0; i < n; i++){
		if(s[i] == 'A' && vis[i][0] == 0 && dfs(i, 0)){
			cout<<"Yes\n";
			exit(0);
		}
	}
	cout<<"No\n";

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

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

const int N = 2e5 + 25;
vector<int> g[N];
int vis[N][4];
string s;

bool dfs(int v, int i){
	if(vis[v][i] == 2)return false;
	if(vis[v][i] == 1)return true;
	vis[v][i] = 1;
	for(int u : g[v]){
		if(s[u] == 'A'){
			if(i == 0 && dfs(u, 1))return true;
			if(i == 3 && dfs(u, 0))return true;
		}else{
			if(i == 1 && dfs(u, 2))return true;
			if(i == 2 && dfs(u, 3))return true;
		}
	}
	vis[v][i] = 2;
	return false;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout<<setprecision(32);

	int n, m;
	cin>>n>>m>>s;
	for(int i = 0; i < m; i++){
		int u, v;
		cin>>u>>v;
		u--; v--;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	memset(vis, 0, sizeof(vis));
	for(int i = 0; i < n; i++){
		if(s[i] == 'A' && vis[i][0] == 0 && dfs(i, 0)){
			cout<<"Yes\n";
			exit(0);
		}
	}
	cout<<"No\n";

	return 0;
}
```