There is a grid with H rows and W columns, where each square is painted black or white.

You are given H strings S_1, S_2, ..., S_H, each of length W. If the square at the i-th row from the top and the j-th column from the left is painted black, the j-th character in the string S_i is `#`; if that square is painted white, the j-th character in the string S_i is `.`.

Find the number of pairs of a black square c_1 and a white square c_2 that satisfy the following condition:

* There is a path from the square c_1 to the square c_2 where we repeatedly move to a vertically or horizontally adjacent square through an alternating sequence of black and white squares: black, white, black, white...

Constraints

* 1 \leq H, W \leq 400
* |S_i| = W (1 \leq i \leq H)
* For each i (1 \leq i \leq H), the string S_i consists of characters `#` and `.`.

Input

Input is given from Standard Input in the following format:


H W
S_1
S_2
:
S_H


Output

Print the answer.

Examples

Input

3 3
.#.
..#
#..


Output

10


Input

3 3
.#.
..#
..


Output

10


Input

2 4
....
....


Output

0


Input

4 3


...


Output

6
import sys
sys.setrecursionlimit(10 ** 7)

H, W = map(int, input().split())
F = [list(input()) for _ in range(H)]

move = [(-1, 0), (1, 0), (0, -1), (0, 1)]
seen = [[0] * W for _ in range(H)]

# Fの点を頂点とするグラフを考える。グラフの連結成分に'#'がbコ, '.'がwコある場合、
# b*wが条件を満たす

def dfs(sy, sx):
    seen[sy][sx] = 1
    b, w = 1, 0
    
    for dy, dx in move:
      ny = sy + dy
      nx = sx + dx
      if 0 <= ny < H and 0 <= nx < W and F[ny][nx] != F[sy][sx] and not seen[ny][nx]:
        p = dfs(ny, nx)
        b += p[1]  # 逆にしている
        w += p[0]  # 天才か？
    return (b, w)
        
count = 0
for i in range(H):
    for j in range(W):
        if not seen[i][j]:
            b, w = dfs(i, j)
            count += b * w

print(count)