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;
typedef long long ll;

bool used[555][555];
ll dx[4]={1,0,-1,0};
ll dy[4]={0,1,0,-1};
ll bl,wh;
ll h,w;
string s[555];
void dfs(ll x,ll y,bool black){
    used[x][y]=1;
    if(black) bl++;
    else wh++;
    for(ll i=0;i<4;i++){
        ll nx=x+dx[i],ny=y+dy[i];
        if(nx<h && nx>=0 && ny>=0 && ny<w && !used[nx][ny]){
            if(black && s[nx][ny]=='.') dfs(nx,ny,0);
            if(!black && s[nx][ny]=='#') dfs(nx,ny,1);
        }
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);


    cin>>h>>w;
    for(ll i=0;i<h;i++){
        cin>>s[i];
    }

    ll ans=0;
    
    for(ll i=0;i<h;i++){
        for(ll j=0;j<w;j++){
            bl=0,wh=0;
            if(!used[i][j] && s[i][j]=='.') dfs(i,j,0);
            if(!used[i][j] && s[i][j]=='#') dfs(i,j,1);
            ans += bl*wh; 
            //cerr<<bl<<" "<<wh<<endl;
        }
    }
    cout<<ans<<endl;


    
}
