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;
vector<string> s;
vector<vector<bool>> visited;
vector<int> dx = {1,0,-1,0};
vector<int> dy = {0,1,0,-1};

long long int black, white;
void DFS(int x, int y){
    if(visited[x][y] == true) return ;

    visited[x][y] = true;
    if(s[x][y] == '#') black += 1;
    if(s[x][y] == '.') white += 1;

    for(int i=0; i<4; i++){
        if(0 <= x+dx[i] && x+dx[i] < h
        && 0 <= y+dy[i] && y+dy[i] < w
        && s[x][y] != s[x+dx[i]][y+dy[i]]){
            DFS(x+dx[i],y+dy[i]);
        }
    }
}

int main(){
    cin >> h >> w;

    s.resize(h);
    for(int i=0; i<h; i++){
        cin >> s[i];
    }

    long long int ans = 0;
    visited.assign(h, vector<bool>(w,false));
    for(int x=0; x<h; x++){
        for(int y=0; y<w; y++){
            if(visited[x][y] == false){
                black = 0;
                white = 0;
                DFS(x,y);
                ans += black * white;
            }
        }
    }
    cout << ans << endl;
    return 0;
}