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 <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#define MOD 1000000007
typedef long long ll;
using namespace std;

string s[410];
int used[410][410];
int h,w;

int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};

ll black,white;

void dfs(int x,int y,int col){
  used[x][y]=col;
  if(s[x][y]=='#') black++;
  else white++;

  for(int i=0;i<4;i++){
    int nx=x+dx[i];
    int ny=y+dy[i];
    if(nx<0||ny<0||nx>=h|ny>=w) continue;
    if(used[nx][ny]!=0) continue;
    if(col==1){
      if(s[nx][ny]=='#') dfs(nx,ny,-col);
    }else{
      if(s[nx][ny]=='.') dfs(nx,ny,-col);
    }
  }
}

int main(){
  //int h,w;
  cin>>h>>w;

  for(int i=0;i<h;i++){
    cin>>s[i];
  }


  ll ans=0;
  for(int i=0;i<h;i++){
    for(int j=0;j<w;j++){
      if(used[i][j]==0&&s[i][j]=='#'){
        dfs(i,j,-1);
        ans=ans+black*white;
        black=0;
        white=0;
      }
    }
  }

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