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;

int n, m;
string s[410];
char used[410][410];
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, -1, 0, 1};

bool in(int i, int j){
    return i >= 0 && i < n && j >= 0 && j < m;
}

int wh, bl;

void dfs(int i, int j){
    used[i][j] = true;
    if(s[i][j] == '#') bl++;
    else wh++;
    for(int k = 0; k < 4; k++){
        int ii = i + dx[k];
        int jj = j + dy[k];
        if(in(ii, jj) && !used[ii][jj] && s[i][j] != s[ii][jj]){
            dfs(ii, jj);
        }
    }
}

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin >> n >> m;
    for(int i = 0; i < n; i++){
        cin >> s[i];
    }
    long long an = 0;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            if(!used[i][j]){
                bl = 0;
                wh = 0;
                dfs(i, j);
                an += 1ll * bl * wh;
            }
        }
    }
    cout << an;
    return 0;
}
