Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int INF = 1e9;
const int MOD = 1e9 + 7;

int h, w;
string s[500];

int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
bool check[500][500];

void dfs(int y, int x, ll &b_num, ll &w_num)
{
  check[y][x] = true;
  for (int i = 0; i < 4; ++i)
  {
    int ny = y + dy[i], nx = x + dx[i];
    if (0 <= nx && nx < w && 0 <= ny && ny < h && !check[ny][nx] && s[ny][nx] != s[y][x])
    {
      if (s[ny][nx] == '#')
        b_num++;
      else
        w_num++;
      dfs(ny, nx, b_num, w_num);
    }
  }
  return;
}

int main()
{
  cin >> h >> w;
  for (int i = 0; i < h; ++i)
    cin >> s[i];

  ll ans = 0;
  for (int i = 0; i < h; ++i)
  {
    for (int j = 0; j < w; ++j)
    {
      if (!check[i][j])
      {
        ll b_num = 0, w_num = 0;
        if (s[i][j] == '#')
          b_num++;
        else
          w_num++;
        dfs(i, j, b_num, w_num);
        ans += b_num * w_num;
      }
    }
  }

  cout << ans << endl;

  return 0;
}