Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

2 3
AB
1 1
1 2
2 2

Sample Output 1

Yes

Since Nusook can visit Vertex 1 and Vertex 2 in any way he likes, he can make all strings consisting of A and B.

Sample Input 2

4 3
ABAB
1 2
2 3
3 1

Sample Output 2

No

For example, Nusook cannot make BB.

The given graph may not be connected.

Sample Input 3

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

Sample Output 3

Yes

Sample Input 4

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

Sample Output 4

No

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::io::Read;
use std::collections::VecDeque;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).ok();
    let mut it = buf.split_whitespace();
    let n = it.next().unwrap().parse::<usize>().unwrap();
    let m = it.next().unwrap().parse::<usize>().unwrap();
    let s = it.next().unwrap().as_bytes();
    let mut to = vec![vec![0; n]; 2];
    let mut g = vec![Vec::new(); n];
    for _ in 0..m {
        let a = it.next().unwrap().parse::<usize>().unwrap() - 1;
        let b = it.next().unwrap().parse::<usize>().unwrap() - 1;
        to[(s[b]-b'A') as usize][a] += 1;
        to[(s[a]-b'A') as usize][b] += 1;
        g[a].push(b);
        g[b].push(a);
    }
    let mut visit = vec![0; n];
    let mut qu = VecDeque::new();
    for i in 0..n {
        if to[0][i] == 0 || to[1][i] == 0 {
            qu.push_back(i);
            visit[i] = 1;
        }
    }
    while !qu.is_empty() {
        let p = qu.pop_front().unwrap();
        let idx = (s[p] - b'A') as usize;
        for &t in &g[p] {
            to[idx][t] -= 1;
            if to[idx][t] == 0 && visit[t] == 0 {
                qu.push_back(t);
                visit[t] = 1;
            }
        }
    }
    println!("{}", if visit.iter().any(|&x| x == 0) { "Yes" } else { "No" });
}