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;
using ll = long long;
using P = pair<ll, ll>;
#define rep(i, m, n) for (int i = m; i < n; ++i)

int vx[4] = {1, 0, -1, 0};
int vy[4] = {0, 1, 0, -1};

int main() {
  int H, W;
  cin >> H >> W;
  vector<string> S(H);
  rep(i, 0, H) cin >> S[i];
  vector<vector<int>> visited(H, vector<int>(W));
  queue<P> Q;
  ll ans = 0;
  rep(i, 0, H) rep(j, 0, W) {
    if (!visited[i][j]) {
      ll h = 0, b = 0;
      visited[i][j] = 1;
      Q.push({i, j});
      while (!Q.empty()) {
        P cur = Q.front(); Q.pop();
        int y = cur.first;
        int x = cur.second;
        if (S[y][x] == '.') h++;
        else b++;
        rep(k, 0, 4) {
          int nx = x + vx[k];
          int ny = y + vy[k];
          if (nx < 0 || W <= nx || ny < 0 || H <= ny) continue;
          if (S[y][x] == S[ny][nx] || visited[ny][nx]) continue;
          visited[ny][nx] = 1;
          Q.push({ny, nx});
        }
      }
      ans += h * b;
    }
  }
  cout << ans << endl;
  return 0;
}
