Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
# define int long long

int H,W;
char E[410][410];
int checked[410][410];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
int blk,wht;
char c; 

void dfs(int x,int y){
  
  checked[x][y]=1;
  if(E[x][y]=='#'){
    blk++;
    }
  else{
    wht++;
  }

  for(int i=0;i<4;i++){
    int ny=y+dy[i];
    int nx=x+dx[i];
    if(nx< 0 || nx >=H || ny<0 || ny>=W || checked[nx][ny]==1) continue;
    if(E[x][y] == E[nx][ny]) continue;
    dfs(nx,ny);
  }
}
 

signed main(){
  cin>>H>>W;
  string s;
  for(int i=0;i<H;i++){
    for(int j=0;j<W;j++){
      checked[i][j]=0;
    }
  }
  
  //入力
  for(int i=0;i<H;i++){
    cin>>s;
    for(int j=0;j<W;j++){
      E[i][j]=s.at(j);
      //cout<<E[i][j];
    }
    //cout<<endl;
  }

  long long ans=0;
  for(int i=0;i<H;i++){
    for(int j=0;j<W;j++){
      if(checked[i][j]==0){
        blk=0;
        wht=0;
        dfs(i,j);
        ans+=blk*wht; 
      }
    }
  }

  cout<<ans<<endl;
  return 0;
}