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 vis[405][405];
char mat[405][405];

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

int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> (mat[i] + 1);
  }
  long long sol = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      if (vis[i][j]) continue;
      int c1 = 0;
      int c2 = 0;
      queue<array<int, 2>> q;
      for (q.push({i, j}); q.size(); q.pop()) {
        int x = q.front()[0];
        int y = q.front()[1];
        if (vis[x][y]) continue;
        vis[x][y] = 1;
        c1 += mat[x][y] == '#';
        c2 += mat[x][y] == '.';
        for (int i = 0; i < 4; ++i) {
          if (mat[x + dx[i]][y + dy[i]]) {
            if (mat[x][y] != mat[x + dx[i]][y + dy[i]]) {
              q.push({x + dx[i], y + dy[i]});
            }
          }
        }
      }
      sol += 1LL * c1 * c2;
    }
  }
  cout << sol << '\n';
}
