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

const int N = 2e5;
set<int> edges[N][2];
int used[N];

void solve() {
	int n, m;
	cin >> n >> m;

	string s;
	cin >> s;
	for (auto &x: s)
	 	x -= 'A';

	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		a--, b--;
		edges[a][s[b]].insert(b);
		edges[b][s[a]].insert(a);
	}

	int cnt = n;

	queue<int> q;
	for (int i = 0; i < n; i++)
		if (edges[i][0].empty() || edges[i][1].empty()) {
			q.push(i);
			used[i] = 1;
			cnt--;
		}

	while (!q.empty()) {
		int v = q.front(); q.pop();
		for (int t = 0; t < 2; t++)
			for (auto u: edges[v][t]) {
				if (used[u])
					continue;
				edges[u][s[v]].erase(v);
				if (edges[u][s[v]].empty()) {
					q.push(u);
					used[u] = 1;
					cnt--;
				}
			}
	}

	cout << (cnt >= 2 ? "Yes" : "No");
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	solve();
}

### 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;

const int N = 2e5;
set<int> edges[N][2];
int used[N];

void solve() {
	int n, m;
	cin >> n >> m;

	string s;
	cin >> s;
	for (auto &x: s)
	 	x -= 'A';

	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		a--, b--;
		edges[a][s[b]].insert(b);
		edges[b][s[a]].insert(a);
	}

	int cnt = n;

	queue<int> q;
	for (int i = 0; i < n; i++)
		if (edges[i][0].empty() || edges[i][1].empty()) {
			q.push(i);
			used[i] = 1;
			cnt--;
		}

	while (!q.empty()) {
		int v = q.front(); q.pop();
		for (int t = 0; t < 2; t++)
			for (auto u: edges[v][t]) {
				if (used[u])
					continue;
				edges[u][s[v]].erase(v);
				if (edges[u][s[v]].empty()) {
					q.push(u);
					used[u] = 1;
					cnt--;
				}
			}
	}

	cout << (cnt >= 2 ? "Yes" : "No");
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	solve();
}

```