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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
using ll = long long;

int h, w;
char s[401][400];
ll a[2];

void fill(int i, int j, char c) {
    if (s[i][j] != c) return;
    a[(i + j) % 2]++;
    s[i][j] = '@';
    if (i > 0) fill(i - 1, j, c);
    if (i < h - 1) fill(i + 1, j, c);
    if (j > 0) fill(i, j - 1, c);
    if (j < w - 1) fill(i, j + 1, c);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> h >> w;

    for (int i = 0; i < h; i++) {
        cin >> s[i];
    }

    for (int i = 0; i < h; i++) {
        for (int j = 0; j < w; j++) {
            if ((i + j) % 2) s[i][j] ^= '#' ^ '.';
        }
    }

    ll r = 0;
    for (int i = 0; i < h; i++) {
        for (int j = 0; j < w; j++) {
            if (s[i][j] == '@') continue;
            a[0] = a[1] = 0;
            fill(i, j, s[i][j]);
            r += a[0] * a[1];
        }
    }

    cout << r << endl;

    return 0;
}