Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
int h, w, dx[] = {0, 0, -1, 1}, dy[] = {1, -1, 0, 0};
long long a, b, ans;
char s[405][405], t;

void dfs(int x, int y) {
    char t = s[x][y];
    if (t == '#') {
        a++;
    } else {
        b++;
    }
    // 同じマスを重複してカウントしないようにする
    s[x][y] = 0;
    for (int i = 0; i < 4; i++) {
        char u = s[x + dx[i]][y + dy[i]];
        // まだ通っていないマス＆次マスの色が異なる
        if (u > 0 && t != u) {
            dfs(x + dx[i], y + dy[i]);
        }
    }
}

int main() {
    cin >> h >> w;
    for (int i = 1; i <= h; i++) {
        cin >> s[i] + 1;
        s[i][w + 1] = 0;
    }
    for (int i = 1; i <= h; i++) {
        for (int j = 1; j <= w; j++) {
            if (s[i][j]) {
                a = b = 0;
                dfs(i, j);
                ans += a * b;
            }
        }
    }
    cout << ans << endl;
    return 0;
}