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;

int v[410][410];
long long ans = 0;
int x[4] = {1, -1, 0, 0};
int y[4] = {0, 0, 1, -1};
vector< vector<bool> > used;
long long c, d;

void search(int i, int j, int vi, int vj) {
  if (v[i][j] == 1) c++;
  else d++;
  used[i][j] = true;
  for (int k = 0; k < 4; k++) {
    if (used[x[k]+i][y[k]+j]) continue;
    if (v[x[k]+i][y[k]+j] == v[i][j] * -1) {
      search(x[k]+i, y[k]+j, vi, vj);
    }
  }
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int h, w; cin >> h >> w;
  for (int i = 0; i < h; i++) {
    string s; cin >> s;
    for (int j = 0; j < w; j++) {
      if (s[j] == '#') v[i+1][j+1] = 1;
      else v[i+1][j+1] = -1;
    }
  }
  used.assign(h + 2, vector<bool>(w + 2, false));
  for (int i = 1; i <= h; i++) {
    for (int j = 1; j <= w; j++) {
      if (used[i][j]) continue;
      c = 0; d = 0;
      search(i, j, i, j);
      ans += c * d;
    }
  }
  cout << ans << endl;
  return 0;
}
