#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define all(a) begin(a), end(a)
#define pb emplace_back

const ll maxn = 5e5 + 5;
ll n, m;
vector<ll> g[maxn];
ll used[maxn];

void dfs(ll i)
{
    if (used[i] == 2) return;
    if (used[i] == 1)
    {
        cout << "Yes" << endl;
        exit(0);
    }
    used[i] = 1;
    for (ll j : g[i])
    {
        dfs(j);
    }
    used[i] = 2;
}

int main()
{
    cin >> n >> m;
    string s;
    s.reserve(maxn);
    cin >> s;
    while (m--)
    {
        ll a, b;
        cin >> a >> b;
        --a; --b;
        if (s[a] == s[b])
        {
            g[a].pb(b + n);
            g[b].pb(a + n);
        }
        else
        {
            g[a + n].pb(b);
            g[b + n].pb(a);
        }
    }
    for (ll i = 0; i < n; ++i)
    {
        if (!used[i])
        {
            dfs(i);
        }
    }
    cout << "No" << endl;
    return 0;
}

### 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;
#define all(a) begin(a), end(a)
#define pb emplace_back

const ll maxn = 5e5 + 5;
ll n, m;
vector<ll> g[maxn];
ll used[maxn];

void dfs(ll i)
{
    if (used[i] == 2) return;
    if (used[i] == 1)
    {
        cout << "Yes" << endl;
        exit(0);
    }
    used[i] = 1;
    for (ll j : g[i])
    {
        dfs(j);
    }
    used[i] = 2;
}

int main()
{
    cin >> n >> m;
    string s;
    s.reserve(maxn);
    cin >> s;
    while (m--)
    {
        ll a, b;
        cin >> a >> b;
        --a; --b;
        if (s[a] == s[b])
        {
            g[a].pb(b + n);
            g[b].pb(a + n);
        }
        else
        {
            g[a + n].pb(b);
            g[b + n].pb(a);
        }
    }
    for (ll i = 0; i < n; ++i)
    {
        if (!used[i])
        {
            dfs(i);
        }
    }
    cout << "No" << endl;
    return 0;
}

```