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
h, w = map(int, input().split())
grid = [[1 if i == "#" else 0 for i in input()] for _ in range(h)]
seen = [[0] * w for _ in range(h)]
DX = (0, 1, 0, -1)
DY = (1, 0, -1, 0)
ans = 0


def dfs(X, Y, cnt):
    todo = [(X, Y)]
    seen[X][Y] = cnt
    black = 0
    white = 1
    while todo:
        x, y = todo.pop()
        now = grid[x][y]
        for dx, dy in zip(DX, DY):
            nx = x + dx
            ny = y + dy
            if nx < 0 or ny < 0 or nx >= h or ny >= w:  # インデックスが範囲外ならだめ
                continue
            if seen[nx][ny] == cnt or now == grid[nx][ny]:  # 既に見ていたり、白-白、黒-黒ならばだめ
                continue
            if 1-now:  # 現在位置が白色の場合
                black += 1
            else:
                white += 1
            seen[nx][ny] = cnt
            todo.append((nx, ny))
    return black*white


cnt = 1
for i in range(h):
    for j in range(w):
        if grid[i][j] or seen[i][j]:
            continue
        ans += dfs(i, j, cnt)
        cnt += 1

print(ans)
