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;

const vector<int> dx = {1, -1, 0, 0}, dy = {0, 0, 1, -1};

int H, W;
vector<string> S;
vector<vector<bool>> seen;

ll cntb = 0, cntw = 0;

void dfs(int y, int x) {
  seen[y][x] = true;
  if (S[y][x] == '#') {
    cntb++;
  } else {
    cntw++;
  }

  for (int i = 0; i < 4; i++) {
    int nx = x + dx[i], ny = y + dy[i];
    if (nx < 0 || nx >= W || ny < 0 || ny >= H) continue;
    if (S[ny][nx] == S[y][x]) continue;
    if (seen[ny][nx]) continue;

    dfs(ny, nx);
  }
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> H >> W;
  S.resize(H), seen.resize(H, vector<bool>(W, false));
  for (int i = 0; i < H; i++) {
    cin >> S[i];
  }

  ll res = 0;
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      /* if (S[i][j] == '.') continue; */
      if (seen[i][j]) continue;
      cntb = cntw = 0;
      dfs(i, j);
      res += cntb * cntw;
    }
  }

  cout << res << '\n';
  return 0;
}
