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

const int N = 2e5 + 10;

int n;
string s;
vector<int> adj[N];
int cnt[N][2], alive[N];

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    int m; cin >> n >> m;
    cin >> s;
    while (m--) {
        int u,v; cin >> u >> v;
        --u; --v;
        adj[u].push_back(v);
        adj[v].push_back(u);
        cnt[u][s[v]-'A']++;
        cnt[v][s[u]-'A']++;
    }
    fill_n(alive, n, 1);
    vector<int> bad;
    for (int u = 0; u < n; u++) if (!cnt[u][0] || !cnt[u][1])
        bad.push_back(u);

    while (!bad.empty()) {
        auto u = bad.back(); bad.pop_back();
        alive[u] = 0;
        for (auto v : adj[u]) if (alive[v]) {
            if (--cnt[v][s[u]-'A'] == 0) bad.push_back(v);
        }
    }
    bool ans = 0;
    for (int u = 0; u < n; u++) ans |= alive[u];
    cout << (ans ? "Yes" : "No");

    return 0;
}

### Prompt

Your task is to create 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 + 10;

int n;
string s;
vector<int> adj[N];
int cnt[N][2], alive[N];

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    int m; cin >> n >> m;
    cin >> s;
    while (m--) {
        int u,v; cin >> u >> v;
        --u; --v;
        adj[u].push_back(v);
        adj[v].push_back(u);
        cnt[u][s[v]-'A']++;
        cnt[v][s[u]-'A']++;
    }
    fill_n(alive, n, 1);
    vector<int> bad;
    for (int u = 0; u < n; u++) if (!cnt[u][0] || !cnt[u][1])
        bad.push_back(u);

    while (!bad.empty()) {
        auto u = bad.back(); bad.pop_back();
        alive[u] = 0;
        for (auto v : adj[u]) if (alive[v]) {
            if (--cnt[v][s[u]-'A'] == 0) bad.push_back(v);
        }
    }
    bool ans = 0;
    for (int u = 0; u < n; u++) ans |= alive[u];
    cout << (ans ? "Yes" : "No");

    return 0;
}

```