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;
typedef long long               ll;
const int mod = 1e9+7;
#ifndef M_PI
const double M_PI = acos(-1.0);
#endif

int h, w;
vector<string> grid;

int cur[2];

void dfs(int i, int j){
  if(grid[i][j] == '#'){
    ++cur[0];
  } else if(grid[i][j] == '.'){
    ++cur[1];
  } else {
    return;
  }
  char old = grid[i][j];
  grid[i][j] = 'u';
  if(i>0 && old != grid[i-1][j])
    dfs(i-1, j);
  if(i<h-1 && old != grid[i+1][j])
    dfs(i+1, j);
  if(j>0 && old != grid[i][j-1])
    dfs(i, j-1);
  if(j<w-1 && old != grid[i][j+1])
    dfs(i, j+1);
}

int main(){
  cin.sync_with_stdio(false);
  cin>>h>>w;
  grid.resize(h);
  for(int i=0; i<h; ++i){
    cin>>grid[i];
  }
  ll res = 0;
  for(int i=0; i<h; ++i){
    for(int j=0; j<w; ++j){
      cur[0] = cur[1] = 0;
      dfs(i, j);
      res += (ll)cur[0]*cur[1];
    }
  }
  cout<<res<<endl;
	return 0;
}
