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;

long long h,w,num1,num2,ans;
char mat[405][405];
bool vis[405][405];

void dfs(int x,int y){
    
    if(mat[x][y]=='.')num1++;
    else num2++;
    
    vis[x][y]=true;
    
    if(x-1>=0 && vis[x-1][y]==false && mat[x-1][y]!=mat[x][y])dfs(x-1,y);
    if(x+1<h && vis[x+1][y]==false && mat[x+1][y]!=mat[x][y])dfs(x+1,y);
    if(y-1>=0 && vis[x][y-1]==false && mat[x][y-1]!=mat[x][y])dfs(x,y-1);
    if(y+1<w && vis[x][y+1]==false && mat[x][y+1]!=mat[x][y])dfs(x,y+1);
}


int main(){

	cin>>h>>w;
	
	for(int i=0;i<h;i++){
	    for(int j=0;j<w;j++){
	        cin>>mat[i][j];
        }
    }
    
    for(int i=0;i<h;i++){
        for(int j=0;j<w;j++){
            if(vis[i][j]==false){
                num1=num2=0;
                dfs(i,j);
                ans+=num1*num2;
                
                //cout<<num1<<" "<<num2<<endl;
            }
        }
    }
    
    cout<<ans<<endl;

return 0;
}
