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
#include<cstdio>
#include<algorithm>
#include<map>

int h,w;
char m[500][500];

void dfs(int y, int x, int &bb, int &ww, char c) {
    if(!(0 <= x && x < w)) return;
    if(!(0 <= y && y < h)) return;
    if(m[y][x] != c) return;
    m[y][x] = '!';
    char ic = c == '#' ? '.' : '#';

    c == '#' ? bb++ : ww++;

    dfs(y+1, x, bb, ww, ic);
    dfs(y-1, x, bb, ww, ic);
    dfs(y, x+1, bb, ww, ic);
    dfs(y, x-1, bb, ww, ic);
    return;
}

int main(void) {
    scanf("%d%d",&h,&w);
    for(int i=0; i<h; i++)
        scanf("%s", m[i]);

    long long int r = 0;
    for(int i=0; i<h; i++)
        for(int j=0; j<w; j++) {
            int bb=0, ww=0;
            dfs(i, j, bb, ww, '.');
            dfs(i, j, bb, ww, '#');
            r += 1ULL*bb*ww;
        }

    printf("%lld\n", r);
}
