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>
using namespace std;
string s[400];
bool done[400][400];
int h, w; 

int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};

void dfs(int i, int j, long long &white, long long &black) {
    done[i][j] = true;
    if (s[i][j] == '.') white++; else black++;
    for (int k = 0; k < 4; k++) {
        int i1 = i + dx[k], j1 = j + dy[k];
        if (0 <= i1 && i1 < h && 0 <= j1 && j1 < w &&
            !done[i1][j1] && s[i][j] != s[i1][j1]) {
            dfs(i1, j1, white, black);
        }
    }
}

int main() {
    cin >> h >> w;
    for (int i = 0; i < h; i++) cin >> s[i];
    long long ans = 0;
    for (int i = 0; i < h; i++)
        for (int j = 0; j < w; j++) {
            if (done[i][j]) continue;
            long long white = 0, black = 0;
            dfs(i, j, white, black);
            ans += white * black;
            //cerr << i << ' ' << j << ' ' << white << ' ' << black << endl;
        }
    cout << ans << endl;
}
