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 rep(i, n) for(int i = 0; i < (int)(n); i++)
typedef long long int ll;
typedef pair<int, int> P;

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

int main()
{
  int h, w;
  cin >> h >> w;
  
  vector<string> s(h);
  rep(i,h) cin >> s[i];
  
  vector<vector<bool>> a(h, vector<bool>(w, false));
  
  ll res = 0;
  rep(i,h) rep(j,w)
  {
    if(a[i][j]) continue;
    
    ll bb = (s[i][j] == '#'), ww = 1 - bb;
    queue<P> q;
    q.push(P(i, j));
    a[i][j] = true;
    
    while(!q.empty())
    {
      auto p = q.front();
      q.pop();
      rep(k,4)
      {
        int ny = p.first + dy[k], nx = p.second + dx[k];
        if(ny < 0 || ny >= h || nx < 0 || nx >= w || a[ny][nx] || 
          s[p.first][p.second] == s[ny][nx]) continue;
        q.push(P(ny, nx));
        a[ny][nx] = true;
        if(s[ny][nx] == '#') bb++;
        else ww++;
      }
    }
    res += bb * ww;
  }
  
  cout << res << endl;
  
  return 0;
}