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>
#include <vector>
using namespace std;

const int MAX = 410;
int done[MAX][MAX];
string s[MAX];

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int h,w;
long long nb, nw;

void dfs(int i, int j){
    // if(done[i][j] > 0) return;
    done[i][j] = 1;
    s[i][j] == '#'? nb++: nw++;
    for(int d=0; d<4; d++){
        int l = i+dx[d];
        int m = j+dy[d];
        if(l<0 || l>=h || m<0 || m>=w) continue;
        if(s[i][j] == s[l][m]) continue;
        if(done[l][m] > 0) continue;
        dfs(l, m);
    }
}

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++){
            nb = 0;
            nw = 0;
            dfs(i, j);
            ans += nb*nw;
            // cout << i << ", "<< j << ": "<< nb << ", " << nw << endl;
        }
    }
    cout << ans << endl; 
    return 0;
}