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;
using lint = long long;

int H, W;
char s[405][405];
bool visited[405][405];
const int dx[] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};

void dfs(int x, int y, lint &black, lint &white) {
    if (visited[x][y]) return;
    visited[x][y] = true;

    if (s[x][y] == '#') {
        black++;
    } else {
        white++;
    }
    for (int i = 0; i < 4; i++) {
        int nx = x + dx[i], ny = y + dy[i];
        if (0 <= nx && nx < H && 0 <= ny && ny < W &&
            s[nx][ny] != s[x][y]) {
            dfs(nx, ny, black, white);
        }
    }
}

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

    lint ans = 0;
    for (int i = 0; i < H; i++) {
        for (int j = 0; j < W; j++) {
            if (!visited[i][j]) {
                lint b = 0, w = 0;
                dfs(i, j, b, w);
                ans += b * w;
            }
        }
    }

    cout << ans << endl;
    return 0;
}
