#include <iostream>
#include <vector>
#define llint long long

using namespace std;

llint N, M;
string s;
vector<llint> G[200005];
llint dega[200005], degb[200005];
bool invalid[200005];

void dfs(int v)
{
	invalid[v] = true;
	//cout << v << endl;
	for(int i = 0; i < G[v].size(); i++){
		int u = G[v][i];
		if(invalid[u]) continue;
		if(s[v] == 'A') dega[u]--;
		else degb[u]--;
		if(dega[u] * degb[u] == 0) dfs(u);
	}
}

int main(void)
{
	cin >> N >> M;
	cin >> s;
	s = "#" + s;
	
	llint u, v;
	for(int i = 0; i < M; i++){
		cin >> u >> v;
		G[u].push_back(v);
		if(s[u] == 'A') dega[v]++;
		else degb[v]++; 
		
		G[v].push_back(u);
		if(s[v] == 'A') dega[u]++;
		else degb[u]++;
	}
	
	/*for(int i = 1; i <= N; i++){
		cout << dega[i] << " " << degb[i] << endl;
	}*/
	
	for(int i = 1; i <= N; i++){
		if(invalid[i]) continue;
		if(dega[i] * degb[i] == 0) dfs(i);
	}
	
	for(int i = 1; i <= N; i++){
		if(!invalid[i]){
			cout << "Yes" << endl;
			return 0;
		}
	}
	cout << "No" << endl;
	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
#include <iostream>
#include <vector>
#define llint long long

using namespace std;

llint N, M;
string s;
vector<llint> G[200005];
llint dega[200005], degb[200005];
bool invalid[200005];

void dfs(int v)
{
	invalid[v] = true;
	//cout << v << endl;
	for(int i = 0; i < G[v].size(); i++){
		int u = G[v][i];
		if(invalid[u]) continue;
		if(s[v] == 'A') dega[u]--;
		else degb[u]--;
		if(dega[u] * degb[u] == 0) dfs(u);
	}
}

int main(void)
{
	cin >> N >> M;
	cin >> s;
	s = "#" + s;
	
	llint u, v;
	for(int i = 0; i < M; i++){
		cin >> u >> v;
		G[u].push_back(v);
		if(s[u] == 'A') dega[v]++;
		else degb[v]++; 
		
		G[v].push_back(u);
		if(s[v] == 'A') dega[u]++;
		else degb[u]++;
	}
	
	/*for(int i = 1; i <= N; i++){
		cout << dega[i] << " " << degb[i] << endl;
	}*/
	
	for(int i = 1; i <= N; i++){
		if(invalid[i]) continue;
		if(dega[i] * degb[i] == 0) dfs(i);
	}
	
	for(int i = 1; i <= N; i++){
		if(!invalid[i]){
			cout << "Yes" << endl;
			return 0;
		}
	}
	cout << "No" << endl;
	return 0;
}
```