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 <algorithm>
#include <string>
using namespace std;
typedef long long ll;
ll ans=0,bl,wh;char s;
int dx[4]={-1,1,0,0},dy[4]={0,0,-1,1};
int a[405][405],b[405][405];
void dfs(int x,int y){
    b[x][y]=1;
    if(a[x][y]==1)bl++;
    else if(a[x][y]==-1)wh++;
    for(int i=0;i<4;i++){
        if((a[x][y]+a[x+dx[i]][y+dy[i]]==0)&&!b[x+dx[i]][y+dy[i]])dfs(x+dx[i],y+dy[i]);
	}
	return;
}

int main(void){
    int h,w;cin>>h>>w;
    for(int i=1;i<=h;i++){
        for(int j=1;j<=w;j++){
            cin>>s;
            if(s=='#')a[i][j]=1;
            else a[i][j]=-1;
        }
    }
    for(int i=1;i<=h;i++){
        for(int j=1;j<=w;j++){
            if(a[i][j]==1){
                bl=0;wh=0;
                dfs(i,j);
                ans+=bl*wh;
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}