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
from collections import deque


def bfs(y,x):
    global s
    global fl
    global h
    global w
    global ans
    dx = [0,1,0,-1]
    dy = [1,0,-1,0]
    count = 1
    black = 0
    d = deque()
    # appendするときに計算する
    d.append([y,x])
    fl[y][x] = 1
    if s[y][x]=='#':
        black+=1
    while len(d):
        p = d.pop()
        for k in range(4):
            if 0<=p[1]+dx[k] <w and 0<=p[0]+dy[k] <h and s[p[0]][p[1]]!=s[p[0]+dy[k]][p[1]+dx[k]]:
                if fl[p[0]+dy[k]][p[1]+dx[k]]:
                    continue
                d.append([p[0]+dy[k],p[1]+dx[k]])
                count += 1
                fl[p[0]+dy[k]][p[1]+dx[k]] = 1
                if s[p[0]+dy[k]][p[1]+dx[k]]=='#':
                    black+=1
    ans += black*(count-black)


h, w = map(int, input().split())
s = [input() for i in range(h)]
fl = [[0 for _ in range(w)] for _ in range(h)]
ans = 0
for i in range(h):
    for j in range(w):
        if not fl[i][j]:
            bfs(i,j)
print(ans)