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 <bits/stdc++.h>

using namespace std;

const int dirx[] = {1, -1, 0, 0};
const int diry[] = {0, 0, 1, -1};

int black, white, h, w;
bool vis[405][405];
char s[405][405];


void dfs(int x, int y) {
    vis[x][y] = 1;

    if (s[x][y] == '#') black++;
    else white++;

    for (int i = 0; i < 4; i++) {
        int newx = x + dirx[i];
        int newy = y + diry[i];
        if (newx < 0 || newx >= h || newy < 0 || newy >= w) continue;
        if (s[x][y] == s[newx][newy]) continue;
        if (vis[newx][newy]) continue;
        dfs(newx, newy);
    }
}

int main() {
    scanf("%d%d", &h, &w);

    for (int i = 0; i < h; i++) scanf("%s", s[i]);

    long long res = 0;

    for (int i = 0; i < h; i++) {
        for (int j = 0; j < w; j++) {
            if (vis[i][j]) continue;
            black = 0;
            white = 0;
            dfs(i, j);
            res += 1LL * black * white;
        }
    }

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