#include"bits/stdc++.h"
using namespace std;
#define int long long
#define con continue
#define rep(i,n) for(int i=0;i<n;i++)
const long long mod = 1000000007;
const long long inf = 1ll<<61;
typedef pair<int, int> P;
typedef pair<P, int> PP;
typedef pair<int, PP> PPP;
struct edge { int to; int to2; int cost; };

vector<int>G[200006];
int ae[200006], be[200006];
int used[200006];
signed main() {
	int n, m; cin >> n >> m;
	string s; cin >> s;
	rep(i, m) {
		int a, b; cin >> a >> b;
		a--; b--;
		G[a].push_back(b);
		if (a - b)G[b].push_back(a);
	}
	queue<int>Q;
	rep(i, n) {
		rep(j, G[i].size()) {
			if (s[G[i][j]] == 'A')ae[i]++;
			else be[i]++;
		}
		if (ae[i] == 0 || be[i] == 0)Q.push(i);
	}
	while (Q.size()) {
		int x = Q.front();
		Q.pop();
		if (used[x])continue;
		used[x]++;
		rep(i, G[x].size()) {
			if (s[x] == 'A')ae[G[x][i]]--;
			else be[G[x][i]]--;
			if (ae[G[x][i]] == 0 || be[G[x][i]] == 0)Q.push(G[x][i]);
		}
	}
	rep(i, n) {
		if (!used[i]) {
			puts("Yes"); return 0;
		}
	}
	puts("No");
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 int long long
#define con continue
#define rep(i,n) for(int i=0;i<n;i++)
const long long mod = 1000000007;
const long long inf = 1ll<<61;
typedef pair<int, int> P;
typedef pair<P, int> PP;
typedef pair<int, PP> PPP;
struct edge { int to; int to2; int cost; };

vector<int>G[200006];
int ae[200006], be[200006];
int used[200006];
signed main() {
	int n, m; cin >> n >> m;
	string s; cin >> s;
	rep(i, m) {
		int a, b; cin >> a >> b;
		a--; b--;
		G[a].push_back(b);
		if (a - b)G[b].push_back(a);
	}
	queue<int>Q;
	rep(i, n) {
		rep(j, G[i].size()) {
			if (s[G[i][j]] == 'A')ae[i]++;
			else be[i]++;
		}
		if (ae[i] == 0 || be[i] == 0)Q.push(i);
	}
	while (Q.size()) {
		int x = Q.front();
		Q.pop();
		if (used[x])continue;
		used[x]++;
		rep(i, G[x].size()) {
			if (s[x] == 'A')ae[G[x][i]]--;
			else be[G[x][i]]--;
			if (ae[G[x][i]] == 0 || be[G[x][i]] == 0)Q.push(G[x][i]);
		}
	}
	rep(i, n) {
		if (!used[i]) {
			puts("Yes"); return 0;
		}
	}
	puts("No");
}
```