#include <bits/stdc++.h>

using namespace std;

#define debug(a) cerr << #a << ": " << a << endl

typedef long long ll;
typedef pair<int, int> ii;

#define x first
#define y second

const int N = 2e5 + 10;

int n, m;
char s[N];
vector<int> adj[N];
int cnt[N][2];
bool vis[N];
queue<int> q;

int main() {
	#ifdef LOCAL
	freopen("in", "r", stdin);
	freopen("out", "w", stdout);
	#endif

	cin >> n >> m;
	scanf("%s", s + 1);

	while (m--) {
		int u, v;
		scanf("%d %d", &u, &v);

		adj[u].push_back(v);
		cnt[u][s[v] - 'A']++;

		adj[v].push_back(u);
		cnt[v][s[u] - 'A']++;
	}

	for (int i = 1; i <= n; i++) {
		if (cnt[i][0] == 0 || cnt[i][1] == 0)
			q.push(i);
	}

	while (!q.empty()) {
		int at = q.front();
		q.pop();

		if (vis[at]) continue;
		vis[at] = true;

		for (int adj_at : adj[at]) if (!vis[adj_at]) {
			cnt[adj_at][s[at] - 'A']--;
			if (cnt[adj_at][s[at] - 'A'] == 0) q.push(adj_at);
		}
	}

	int visited = accumulate(vis + 1, vis + n + 1, 0);
	if (visited < n) printf("Yes\n");
	else printf("No\n");

	return 0;
}
### Prompt

Please formulate 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;

#define debug(a) cerr << #a << ": " << a << endl

typedef long long ll;
typedef pair<int, int> ii;

#define x first
#define y second

const int N = 2e5 + 10;

int n, m;
char s[N];
vector<int> adj[N];
int cnt[N][2];
bool vis[N];
queue<int> q;

int main() {
	#ifdef LOCAL
	freopen("in", "r", stdin);
	freopen("out", "w", stdout);
	#endif

	cin >> n >> m;
	scanf("%s", s + 1);

	while (m--) {
		int u, v;
		scanf("%d %d", &u, &v);

		adj[u].push_back(v);
		cnt[u][s[v] - 'A']++;

		adj[v].push_back(u);
		cnt[v][s[u] - 'A']++;
	}

	for (int i = 1; i <= n; i++) {
		if (cnt[i][0] == 0 || cnt[i][1] == 0)
			q.push(i);
	}

	while (!q.empty()) {
		int at = q.front();
		q.pop();

		if (vis[at]) continue;
		vis[at] = true;

		for (int adj_at : adj[at]) if (!vis[adj_at]) {
			cnt[adj_at][s[at] - 'A']--;
			if (cnt[adj_at][s[at] - 'A'] == 0) q.push(adj_at);
		}
	}

	int visited = accumulate(vis + 1, vis + n + 1, 0);
	if (visited < n) printf("Yes\n");
	else printf("No\n");

	return 0;
}
```