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: Wrong Answer)

#[allow(unused_imports)]
use std::cmp::{min,max};
#[allow(unused_imports)]
use std::collections::{BTreeMap,BTreeSet};
#[allow(unused_imports)]
use std::ops::*;
#[allow(unused_imports)]
use std::collections::BinaryHeap;

#[allow(unused_macros)]
macro_rules! tf {
    ($c:expr, $t:expr, $f:expr) => {{
        if $c { $t } else { $f }
    }};
}

fn main() {
    let (n,m) = {
        let xs = read_vec_i64();
        (xs[0] as usize, xs[1] as usize)
    };
    let ss: Vec<char> = read_line().chars().collect();
    let edges = {
        let mut edges = vec![];
        for _ in 0..m {
            let xs = read_vec_i64();
            edges.push(((xs[0]-1) as usize, (xs[1]-1) as usize));
        }
        edges
    };
    let mut code = vec![0; n];
    for &(s,t) in &edges {
        if ss[s] == 'A' && ss[t] == 'A' {
            code[s] |= 0b0001;
            code[t] |= 0b0001;
        }
    }
    for &(s,t) in &edges {
        if ss[s] == 'A' && code[s] & 0b0001 != 0 && ss[t] == 'B' {
            code[t] |= 0b0010;
        }
        let (s,t) = (t,s);
        if ss[s] == 'A' && code[s] & 0b0001 != 0 && ss[t] == 'B' {
            code[t] |= 0b0010;
        }
    }
    for &(s,t) in &edges {
        if code[s] & 0b0010 != 0 && code[t] & 0b0010 != 0 {
            println!("Yes");
            return;
        }
    }
    println!("No");
}

#[allow(dead_code)]
fn read_line() -> String {
    let mut ret = String::new();
    std::io::stdin().read_line(&mut ret).ok();
    ret.pop();
    //ret.pop();
    return ret;
}

#[allow(dead_code)]
fn read_i64() -> i64 {
    let ss = read_line();
    return ss.parse::<i64>().unwrap();
}

#[allow(dead_code)]
fn read_vec_i64() -> Vec<i64> {
    let mut res = vec![];
    let ss = read_line();
    for ts in ss.split_whitespace() {
        let x = ts.parse::<i64>().unwrap();
        res.push(x);
    }
    return res;
}

use std::fmt::Display;
#[allow(dead_code)]
fn write_vec<T: Display>(xs: &Vec<T>) {
    if xs.len() == 0 {
        println!("");
        return;
    }
    print!("{}", xs[0]);
    for i in 1..xs.len() {
        print!(" {}", xs[i]);
    }
    println!("");
}