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

const int MAXN = 2e5 + 5;

int N, M;
char s[MAXN];
vector <int> adj[2 * MAXN];
int bio[2 * MAXN];

void add(int u, int v) {
	adj[u].push_back(v);
}

void load() {
	scanf("%d%d%s", &N, &M, s + 1);
	while (M--) {
		int a, b;
		scanf("%d%d", &a, &b);
		if (s[a] == s[b]) {
			add(2 * a - 1, 2 * b);
			add(2 * b - 1, 2 * a);
		}
		else {
			add(2 * a, 2 * b - 1);
			add(2 * b, 2 * a - 1);
		}
	}
}

bool dfs(int x) {
	if (bio[x])
		return 2 - bio[x];
	bio[x] = 1;
	for (auto it : adj[x])
		if (dfs(it))
			return true;
	bio[x] = 2;
	return false;
}

bool solve() {
	for (int i = 1; i <= 2 * N; i++)
		if (dfs(i))
			return true;
	return false;
}

int main() {
	load();
	puts(solve() ? "Yes" : "No");
	return 0;
}
### 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;

const int MAXN = 2e5 + 5;

int N, M;
char s[MAXN];
vector <int> adj[2 * MAXN];
int bio[2 * MAXN];

void add(int u, int v) {
	adj[u].push_back(v);
}

void load() {
	scanf("%d%d%s", &N, &M, s + 1);
	while (M--) {
		int a, b;
		scanf("%d%d", &a, &b);
		if (s[a] == s[b]) {
			add(2 * a - 1, 2 * b);
			add(2 * b - 1, 2 * a);
		}
		else {
			add(2 * a, 2 * b - 1);
			add(2 * b, 2 * a - 1);
		}
	}
}

bool dfs(int x) {
	if (bio[x])
		return 2 - bio[x];
	bio[x] = 1;
	for (auto it : adj[x])
		if (dfs(it))
			return true;
	bio[x] = 2;
	return false;
}

bool solve() {
	for (int i = 1; i <= 2 * N; i++)
		if (dfs(i))
			return true;
	return false;
}

int main() {
	load();
	puts(solve() ? "Yes" : "No");
	return 0;
}
```