#include <bits/stdc++.h>
typedef long long i64;
using std::cout;
using std::endl;
using std::cin;

std::vector<std::vector<int>> g;
std::string s;
int n, m, now = 0;

int main(){
	cin >> n >> m >> s; std::vector<int> a(m), b(m);
	std::vector<std::vector<int>> h(n, std::vector<int>(2, 0)); g.resize(n);
	for(int i = 0; i < m; i++) {
		cin >> a[i] >> b[i]; a[i]--; b[i]--;
		
		g[a[i]].push_back(b[i]);
		g[b[i]].push_back(a[i]);
		h[a[i]][s[b[i]] != 'A']++;
		h[b[i]][s[a[i]] != 'A']++;
	}
	
	std::queue<int> qu;
	std::vector<bool> used(n, false);
	for(int i = 0; i < n; i++) {
		if(!h[i][0] or !h[i][1]) {
			qu.push(i);
			used[i] = true;
		}
	}
	
	while(!qu.empty()) {
		auto p = qu.front(); qu.pop();
		
		for(auto e : g[p]) {
			if(used[e]) continue;
			h[e][s[p] != 'A']--;
			
			if(!h[e][0] or !h[e][1]) {
				qu.push(e);
				used[e] = true;
			}
		}
	}
	
	for(int i = 0; i < n; i++) if(h[i][0] and h[i][1]) {
		cout << "Yes" << endl;
		return 0;
	}
	cout << "No" << 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 <bits/stdc++.h>
typedef long long i64;
using std::cout;
using std::endl;
using std::cin;

std::vector<std::vector<int>> g;
std::string s;
int n, m, now = 0;

int main(){
	cin >> n >> m >> s; std::vector<int> a(m), b(m);
	std::vector<std::vector<int>> h(n, std::vector<int>(2, 0)); g.resize(n);
	for(int i = 0; i < m; i++) {
		cin >> a[i] >> b[i]; a[i]--; b[i]--;
		
		g[a[i]].push_back(b[i]);
		g[b[i]].push_back(a[i]);
		h[a[i]][s[b[i]] != 'A']++;
		h[b[i]][s[a[i]] != 'A']++;
	}
	
	std::queue<int> qu;
	std::vector<bool> used(n, false);
	for(int i = 0; i < n; i++) {
		if(!h[i][0] or !h[i][1]) {
			qu.push(i);
			used[i] = true;
		}
	}
	
	while(!qu.empty()) {
		auto p = qu.front(); qu.pop();
		
		for(auto e : g[p]) {
			if(used[e]) continue;
			h[e][s[p] != 'A']--;
			
			if(!h[e][0] or !h[e][1]) {
				qu.push(e);
				used[e] = true;
			}
		}
	}
	
	for(int i = 0; i < n; i++) if(h[i][0] and h[i][1]) {
		cout << "Yes" << endl;
		return 0;
	}
	cout << "No" << endl;
	return 0;
}

```