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)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn main() {
    let v = read_vec::<usize>();
    let (n, m) = (v[0], v[1]);
    let s = read::<String>().chars().collect::<Vec<_>>();
    let mut g = vec![vec![]; n];

    for _ in 0..m {
        let v = read_vec::<usize>();
        let (a, b) = (v[0] - 1, v[1] - 1);
        g[a].push(b);
        g[b].push(a);
    }

    let mut a_count = vec![0; n];
    let mut b_count = vec![0; n];
    for from in 0..n {
        for &to in g[from].iter() {
            if s[from] == 'A' {
                a_count[to] += 1;
            } else {
                b_count[to] += 1;
            }
        }
    }

    let mut new_unuse = HashSet::new();
    let mut usable = vec![true; n];
    for from in 0..n {
        if a_count[from] == 0 || b_count[from] == 0 {
            usable[from] = false;
            new_unuse.insert(from);
        }
    }

    while !new_unuse.is_empty() {
        let mut unuse = HashSet::new();
        std::mem::swap(&mut unuse, &mut new_unuse);
        for &i in unuse.iter() {
            for &to in g[i].iter() {
                if s[i] == 'A' {
                    a_count[to] -= 1;
                } else {
                    b_count[to] -= 1;
                }
                if (a_count[to] == 0 || b_count[to] == 0) && usable[to] {
                    new_unuse.insert(to);
                    usable[to] = false;
                }
            }
        }
    }
    if usable.iter().any(|&x| x) {
        println!("Yes");
    } else {
        println!("No");
    }
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}