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;
ll par[160005],c[160005];
ll a[160005],b[160005];
void init(int x){
  for(int i=0;i<x;i++){
    par[i]=i;
    c[i]=1;
  }
}
ll root(int x){
  if(par[x]==x){
    return x;
  }
  return par[x]=root(par[x]);
}
void unite(int x,int y){
  x=root(x);
  y=root(y);
  if(x==y){
    return;
  }
  par[x]=y;
  c[y]+=c[x];
  b[y]+=b[x];
}

int main(){
  int h,w;cin >> h >> w;
  vector<string> s(h);
  init(h*w);
  for(int i=0;i<h;i++){
    cin >> s[i];
  }
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      if(s[i][j]=='#'){
        b[i*w+j]=1;
      }
    }
  }
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      if(i!=h-1){
        if(s[i][j]!=s[i+1][j]){
          unite(i*w+j,(i+1)*w+j);
        }
      }
      if(j!=w-1){
        if(s[i][j]!=s[i][j+1]){
          unite(i*w+j,i*w+j+1);
        }
      }
    }
  }
  ll ans=0;
  for(int i=0;i<h*w;i++){
    if(root(i)==i){
      ans+=b[i]*(c[i]-b[i]);
    }
  }
  cout << ans << endl;
}