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 <vector>
#include <string.h>
#include <set>
#include <algorithm>
using namespace std;

long long int dx[4] ={1,0,-1,0};
long long int dy[4] = {0,1,0,-1};
long long int cntw,cntb;
long long int state[402][402];

long long int dfs(int x,int y){
  if(state[x][y]==0) {
    cntw+=1;
  }
  if(state[x][y]==1){
    cntb+=1;
  }
  long long int tmp=state[x][y];
  state[x][y]=-1;  
  for(long long int i=0;i<4;i++){
    long long int nx=x+dx[i], ny=y+dy[i];
    if(state[nx][ny]+tmp==1) {
      dfs(nx,ny);
    } 
  }
  return cntb*cntw;
}

int main()
{
  long long int H,W;
  cin>>H>>W;
  memset(state,-1,sizeof(state));

  for(int i=1;i<=H;i++){
    string s;
    cin>>s;
    for(int j=1;j<=W;j++){
      if(s[j-1]=='.'){
        state[i][j]=0;
      }else{
        state[i][j]=1;
      }
    }
  }
  
  long long int ans=0;
  for(int i=0;i<402;i++){
    for(int j=0;j<402;j++){
      cntw=0,cntb=0;
      if(state[i][j]>=0){
	ans += dfs(i,j);
      }
    }
  }
  cout<< ans <<endl;
}
