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

using vi = vector<int>;
#define f(i,a,b) for (int i = a; i < b; i++)
#define pb push_back

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	clock_t start = clock();
#endif

	int n,m;
	cin>>n>>m;
	string s; cin>>s;
	vi adj[n];
	int p[n][2];
	f(i,0,n) p[i][0] = p[i][1] = 0;
	f(i,0,m) {
		int a,b;
		cin>>a>>b;
		adj[--a].pb(--b);
		adj[b].pb(a);
		p[a][s[b]-'A']++;
		p[b][s[a]-'A']++;
	}
	vi pq;
	vi marked(n,0);
	f(i,0,n) if (!p[i][0] || !p[i][1]) {
		pq.pb(i);
		marked[i] = 1;
	}
	while (pq.size()) {
		int v = pq.back(); pq.pop_back();	
		for (int w : adj[v]) if (!marked[w] && !--p[w][s[v]-'A']) {
			pq.pb(w);
			marked[w] = 1;
		}
	}
	bool found = 0;
	f(i,0,n) if (!marked[i]) found = 1;
	cout << (found?"Yes":"No") << endl;

#ifdef LOCAL
	cout << setprecision(12) << (long double)(clock()-start) / CLOCKS_PER_SEC << endl;
#endif
	
	return 0;
}


### Prompt

Your task is to create 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;

using vi = vector<int>;
#define f(i,a,b) for (int i = a; i < b; i++)
#define pb push_back

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	clock_t start = clock();
#endif

	int n,m;
	cin>>n>>m;
	string s; cin>>s;
	vi adj[n];
	int p[n][2];
	f(i,0,n) p[i][0] = p[i][1] = 0;
	f(i,0,m) {
		int a,b;
		cin>>a>>b;
		adj[--a].pb(--b);
		adj[b].pb(a);
		p[a][s[b]-'A']++;
		p[b][s[a]-'A']++;
	}
	vi pq;
	vi marked(n,0);
	f(i,0,n) if (!p[i][0] || !p[i][1]) {
		pq.pb(i);
		marked[i] = 1;
	}
	while (pq.size()) {
		int v = pq.back(); pq.pop_back();	
		for (int w : adj[v]) if (!marked[w] && !--p[w][s[v]-'A']) {
			pq.pb(w);
			marked[w] = 1;
		}
	}
	bool found = 0;
	f(i,0,n) if (!marked[i]) found = 1;
	cout << (found?"Yes":"No") << endl;

#ifdef LOCAL
	cout << setprecision(12) << (long double)(clock()-start) / CLOCKS_PER_SEC << endl;
#endif
	
	return 0;
}


```