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>
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;

int h,w,par[160000],Size[160000][2];
char c[400][400];

int root(int x){
    if(par[x]==x)return x;
    return par[x]=root(par[x]);
}

signed main(){
    cin>>h>>w;
    rep(i,h)rep(j,w){
        cin>>c[i][j];
        par[i*w+j]=i*w+j;
        Size[i*w+j][c[i][j]=='#']=1;
        if(i&&c[i-1][j]!=c[i][j]&&root(i*w+j)!=root(i*w+j-w)){
            int x=root(i*w+j),y=root(i*w+j-w);
            par[x]=y;
            Size[y][0]+=Size[x][0];
            Size[y][1]+=Size[x][1];
            Size[x][0]=Size[x][1]=0;
        }if(j&&c[i][j-1]!=c[i][j]&&root(i*w+j)!=root(i*w+j-1)){
            int x=root(i*w+j),y=root(i*w+j-1);
            par[x]=y;
            Size[y][0]+=Size[x][0];
            Size[y][1]+=Size[x][1];
            Size[x][0]=Size[x][1]=0;
        }
    }int ans=0;
    rep(i,w*h){
        if(par[i]==i){
            ans+=Size[i][0]*Size[i][1];
        }
    }
    cout<<ans<<endl;
}