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 dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int used[405][405];

int main() {
  int H, W;
  cin >> H >> W;
  vector<string> S(H);
  for (int i = 0; i < H; i++) cin >> S[i];

  long ans = 0;
  for (int h = 0; h < H; h++) {
    for (int w = 0; w < W; w++) {
      if (used[h][w]) continue;
      used[h][w] = 1;

      long nb = 0, nw = 0;
      queue<pair<int, int>> Q;
      Q.push(make_pair(h, w));
      while (Q.size()) {
        auto p = Q.front(); Q.pop();
        int h1 = p.first, w1 = p.second;
        if (S[h1][w1] == '.') nw++;
        else nb++;
        for (int i = 0; i < 4; i++) {
          int h2 = h1 + dy[i], w2 = w1 + dx[i];
          if (h2 < 0 || h2 == H || w2 < 0 || w2 == W || S[h2][w2] == S[h1][w1] || used[h2][w2]) continue;
          Q.push(make_pair(h2, w2));
          used[h2][w2] = 1;
        }
      }
      ans += nb * nw;
    }
  }
  cout << ans << endl;
}