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;long a,b;long sum=0;
bool S[400][400];bool image;int check[400][400];
void search(int x,int y){
    if(x<0||x>=H||y<0||y>=W){
    }else if(check[x][y]==1){
    }else if(S[x][y]==image){
        check[x][y]=1;
        if(S[x][y]==true){
            a++;
        }else{
            b++;
        }image=!image;
        search(x-1,y);
        search(x+1,y);
        search(x,y-1);
        search(x,y+1);
        image=!image;
    }return;
}
int main(){
    cin>>H>>W;char temp;
    for(int i=0;i<H;i++){
        for(int j=0;j<W;j++){
            cin>>temp;
            if(temp=='#'){
                S[i][j]=true;
            }else{
                S[i][j]=false;
            }check[i][j]=0;
        }
    }for(int i=0;i<H;i++){
        for(int j=0;j<W;j++){
            if(check[i][j]==0){
                a=0;b=0;
                image=S[i][j];
                search(i,j);
                sum+=a*b;
           }
        }
    }cout<<sum<<endl;
}