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<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<algorithm>
#include<utility>
#include<math.h>

using namespace std;
typedef pair<int, int> P;
typedef long long int ll;

const int INF = 1<<30;

char field[410][410];
int h, w;
int dx[]={0, 0, 1, -1}, dy[]={1, -1, 0, 0};
P voyage(int i, int j){
  int sharp=0, dot=0;
  char c=(field[i][j]=='#'?'.':'#');
  if(field[i][j]=='#')sharp++;
  else dot++;
  field[i][j]='x';
  for(int k=0;k<4;k++){
    int x=i+dx[k], y=j+dy[k];
    if(0<=x&&x<h&&0<=y&&y<w&&field[x][y]==c){
      P p = voyage(x, y);
      sharp+=p.first;dot+=p.second;
    }
  }
  return P(sharp, dot);
}

int main(){
  cin>>h>>w;
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      cin>>field[i][j];
    }
  }
  ll ans=0;
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      if(field[i][j]=='#'){
        P p = voyage(i, j);
        ans+=ll(p.first)*ll(p.second);
      }
    }
  }
  cout<<ans<<endl;
}