#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;
const double EPS = 1e-10;
vector<int> G[400000];
int vis[400000];
bool dfs(int v) {
	if (vis[v] == 1) return 0;
	if (vis[v] == 2) return 1;
	vis[v] = 1;
	for (int to : G[v]) {
		if (!dfs(to)) return 0;
	}
	vis[v] = 2;
	return 1;
}
int main() {
	int N, M;
	cin >> N >> M;
	string s;
	cin >> s;
	for (int i = 0; i < M; i++) {
		int a, b;
		cin >> a >> b;
		a--; b--;
		if (s[a] == s[b]) {
			G[a * 2].push_back(b * 2 + 1);
			G[b * 2].push_back(a * 2 + 1);
		}
		else {
			G[a * 2 + 1].push_back(b * 2);
			G[b * 2 + 1].push_back(a * 2);
		}
	}
	bool ok = 0;
	for (int i = 0; i < N; i++) {
		if (vis[i]) continue;
		if (!dfs(i)) {
			ok = 1;
			break;
		}
	}
	if (ok) {
		cout << "Yes" << endl;
	}
	else {
		cout << "No" << endl;
	}
}
### 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 "bits/stdc++.h"
using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;
const double EPS = 1e-10;
vector<int> G[400000];
int vis[400000];
bool dfs(int v) {
	if (vis[v] == 1) return 0;
	if (vis[v] == 2) return 1;
	vis[v] = 1;
	for (int to : G[v]) {
		if (!dfs(to)) return 0;
	}
	vis[v] = 2;
	return 1;
}
int main() {
	int N, M;
	cin >> N >> M;
	string s;
	cin >> s;
	for (int i = 0; i < M; i++) {
		int a, b;
		cin >> a >> b;
		a--; b--;
		if (s[a] == s[b]) {
			G[a * 2].push_back(b * 2 + 1);
			G[b * 2].push_back(a * 2 + 1);
		}
		else {
			G[a * 2 + 1].push_back(b * 2);
			G[b * 2 + 1].push_back(a * 2);
		}
	}
	bool ok = 0;
	for (int i = 0; i < N; i++) {
		if (vis[i]) continue;
		if (!dfs(i)) {
			ok = 1;
			break;
		}
	}
	if (ok) {
		cout << "Yes" << endl;
	}
	else {
		cout << "No" << endl;
	}
}
```