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 <string>
using namespace std;

typedef pair<long long, long long> P;

int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
int h,w;
string s[400];
bool b[400][400];

P f(int y, int x){
    long long c=0,d=0;
    b[y][x]=true;
    if(s[y][x]=='#'){
        c++;
    }else{
        d++;
    }
    for(int i=0;i<4;i++){
        int nx=x+dx[i],ny=y+dy[i];
        if(ny>=0&&ny<h&&nx>=0&&nx<w){
            if(!b[ny][nx]&&s[y][x]!=s[ny][nx]){
                P p=f(ny,nx);
                c+=p.first;
                d+=p.second;
            }
        }
    }
    return P(c,d);
}

int main(void){
    long long t=0;
    cin>>h>>w;
    for(int i=0;i<h;i++){
        cin>>s[i];
        for(int j=0;j<w;j++){
            b[i][j]=false;
        }
    }
    
    for(int i=0;i<h;i++){
        for(int j=0;j<w;j++){
            if(!b[i][j]){
                P p=f(i, j);
                t+=p.first*p.second;
            }
        }
    }
    
    cout<<t<<endl;
}
