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 H, W;
int one, two;

string S[404];
int visit[404][404];

void dfs(int i, int j, char x)
{
    if(i < 0 || j < 0 || i == H || j == W || S[i][j] != x || visit[i][j])
        return;
    
    visit[i][j] = 1;
    one += (S[i][j] == '#');
    two += (S[i][j] == '.');
    
    dfs(i-1, j, (x == '#' ? '.' : '#'));
    dfs(i+1, j, (x == '#' ? '.' : '#'));
    dfs(i, j-1, (x == '#' ? '.' : '#'));
    dfs(i, j+1, (x == '#' ? '.' : '#'));
}

int main() {
    
    ios::sync_with_stdio(0);
    cin.tie(NULL), cout.tie(NULL);
    
    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(visit[i][j])
                continue;
            
            one = 0, two = 0;
            dfs(i, j, S[i][j]);
            
            ans += 1LL * one * two;
        }
    }
    
    cout << ans << "\n";
    return 0;
}