#include <bits/stdc++.h>
using namespace std;
#define pii pair<int,int>
#define F first
#define S second
#define mt make_tuple
#define pb push_back


string s;
int n, m;
set<int> c[200005][2];
bool deleted[200005];
vector<int> adj[200005];
int32_t main(){
	cin>>n>>m>>s;
	for(int i = 0 ; i < m ; i ++){
		int x , y;
		cin>>x>>y;
		x-- , y--;
		adj[x].pb(y);
		adj[y].pb(x);
		c[x][s[y] - 'A'].insert(y);
		c[y][s[x] - 'A'].insert(x);
	}
	queue<int> q;
	for(int i = 0 ; i < n; i ++){
		if(c[i][0].size() > 0 && c[i][1].size() > 0){
			continue;
		}
		q.push(i);
	}
	bool can = true;
	while(!q.empty()){
		int u = q.front();
		q.pop();
		deleted[u] = true;
		for(auto p : adj[u]){
			if(!deleted[p]){
				c[p][s[u] - 'A'].erase(u);
				if(c[p][0].size() > 0 && c[p][1].size() > 0){
					continue;
				}
				q.push(p);
			}
		}
	}
	can = false;
	for(int i = 0 ; i < n ; i++){
		if(!deleted[i]) can = true;
	}
	if(can) cout<<"Yes\n";
	else cout<<"No\n";
}
### Prompt

Develop a solution in cpp to the problem described below:

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 pii pair<int,int>
#define F first
#define S second
#define mt make_tuple
#define pb push_back


string s;
int n, m;
set<int> c[200005][2];
bool deleted[200005];
vector<int> adj[200005];
int32_t main(){
	cin>>n>>m>>s;
	for(int i = 0 ; i < m ; i ++){
		int x , y;
		cin>>x>>y;
		x-- , y--;
		adj[x].pb(y);
		adj[y].pb(x);
		c[x][s[y] - 'A'].insert(y);
		c[y][s[x] - 'A'].insert(x);
	}
	queue<int> q;
	for(int i = 0 ; i < n; i ++){
		if(c[i][0].size() > 0 && c[i][1].size() > 0){
			continue;
		}
		q.push(i);
	}
	bool can = true;
	while(!q.empty()){
		int u = q.front();
		q.pop();
		deleted[u] = true;
		for(auto p : adj[u]){
			if(!deleted[p]){
				c[p][s[u] - 'A'].erase(u);
				if(c[p][0].size() > 0 && c[p][1].size() > 0){
					continue;
				}
				q.push(p);
			}
		}
	}
	can = false;
	for(int i = 0 ; i < n ; i++){
		if(!deleted[i]) can = true;
	}
	if(can) cout<<"Yes\n";
	else cout<<"No\n";
}
```