p03256 AtCoder Grand Contest 027 - ABland Yard
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
```python
import sys,time
sys.setrecursionlimit(10**7)

start_time = time.time()
N,M = map(int,input().split())
S = input()
src = [tuple(map(lambda x:int(x)-1,sys.stdin.readline().split())) for i in range(M)]

outdeg = [set() for i in range(2*N)]
for x,y in src:
    if S[x] == S[y]:
        #A0->A1, B0->B1
        outdeg[x].add(y+N)
        outdeg[y].add(x+N)
    else:
        #A1->B0, B1->A0
        outdeg[x+N].add(y)
        outdeg[y+N].add(x)

mem = [0] * (2*N)
def visit(v):
    if time.time() - start_time > 1.8:
        # gori~~~
        print('No')
        exit()
    if mem[v] == 1:
        print('Yes')
        exit()
    if mem[v] == 0:
        mem[v] = 1
        for to in outdeg[v]:
            visit(to)
        mem[v] = 2

for i in range(2*N):
    visit(i)

print('No')
```