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 PyPy3 (2.4.0) (judge verdict: Accepted)

import os
import sys
from collections import defaultdict

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353


N, M = list(map(int, sys.stdin.buffer.readline().split()))
S = sys.stdin.buffer.readline().decode().rstrip()
AB = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(M)]

graph = [defaultdict(int) for _ in range(N)]
for a, b in AB:
    a -= 1
    b -= 1
    graph[a][b] += 1
    if a != b:
        graph[b][a] += 1

# AB 両方と隣接する頂点のみで部分グラフを作れるか
# どっちかとしか隣接しない頂点を取り除いていく
# removed = True
# while removed:
#     removed = False
#     for v in range(N):
#         if not graph[v]:
#             continue
#         a = 0
#         b = 0
#         for u in graph[v]:
#             if S[u] == 'A':
#                 a += 1
#             else:
#                 b += 1
#         if a == 0 or b == 0:
#             for u in graph[v]:
#                 graph[u].remove(v)
#             graph[v] = []
#             removed = True
# print(graph)


a_counts = [[0, v] for v in range(N)]
b_counts = [[0, v] for v in range(N)]
for v in range(N):
    for u, cnt in graph[v].items():
        if S[u] == 'A':
            a_counts[v][0] += cnt
        else:
            b_counts[v][0] += cnt
candidates = []
for c, v in a_counts + b_counts:
    if c == 0:
        candidates.append(v)

while candidates:
    v = candidates.pop()
    if not graph[v]:
        continue
    for u, cnt in list(graph[v].items()):
        if S[v] == 'A':
            a_counts[u][0] -= cnt
            if a_counts[u][0] == 0:
                candidates.append(u)
        else:
            b_counts[u][0] -= cnt
            if b_counts[u][0] == 0:
                candidates.append(u)
        del graph[u][v]
    graph[v] = {}

# print(graph)

ok = any(graph)
if ok:
    print('Yes')
else:
    print('No')