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;
#define ll long long
#define mod 1000000007
ll h,w,cnt,a,d;
int x[4]={-1,1,0,0};
int y[4]={0,0,-1,1};
vector<vector<char>> c(402,vector<char>(402,'a'));
bool b[402][402];
set<pair<int,int>> s[402*402];
void dfs(int i,int j){
  b[i][j]=false;
  if(c[i][j]=='#'){
  a++;
    for(int k=0;k<4;k++){
      if(c[i+x[k]][j+y[k]]=='.'&&b[i+x[k]][j+y[k]]){
        dfs(i+x[k],j+y[k]);
      }
    }
  }else if(c[i][j]=='.'){
    s[d].insert(make_pair(i,j));
    for(int k=0;k<4;k++){
      if(c[i+x[k]][j+y[k]]=='#'&&b[i+x[k]][j+y[k]]){
        dfs(i+x[k],j+y[k]);
      }
    }
  }
}
int main(){
  cin>>h>>w;
  for(int i=1;i<=h;i++){
    for(int j=1;j<=w;j++){
      cin>>c[i][j];
    }
  }
  memset(b,true,sizeof(b));
  for(int i=1;i<=h;i++){
    for(int j=1;j<=w;j++){
      if(c[i][j]=='#'){
        dfs(i,j);
        cnt+=(int)s[d].size()*a;
        d++;
        a=0;
      }
    }
  }
  cout<<cnt;
}