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 Lua (LuaJIT 2.1.0) (judge verdict: Accepted)

local n, m = io.read("*n", "*n", "*l")
local s = io.read()
local v = {}
local edge = {}
local acnt, bcnt = {}, {}
for i = 1, n do
  v[i] = s:byte(i) - 64
  edge[i] = {}
  acnt[i], bcnt[i] = 0, 0
end
for i = 1, m do
  local a, b = io.read("*n", "*n")
  table.insert(edge[a], b)
  table.insert(edge[b], a)
  if v[a] == 1 then
    acnt[b] = acnt[b] + 1
  else
    bcnt[b] = bcnt[b] + 1
  end
  if v[b] == 1 then
    acnt[a] = acnt[a] + 1
  else
    bcnt[a] = bcnt[a] + 1
  end
end
local tasks = {}
for i = 1, n do
  if acnt[i] * bcnt[i] == 0 then
    table.insert(tasks, i)
  end
end
local done = 0
while done < #tasks do
  done = done + 1
  local src = tasks[done]
  local cnttbl = v[src] == 1 and acnt or bcnt
  for j = 1, #edge[src] do
    local dst = edge[src][j]
    if 0 < acnt[dst] and 0 < bcnt[dst] then
      cnttbl[dst] = cnttbl[dst] - 1
      if cnttbl[dst] == 0 then
        table.insert(tasks, dst)
      end
    end
  end
end
print(#tasks == n and "No" or "Yes")