Problem:
Score : 300 points

Problem Statement

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.

Sample Input 1

3 3
.#.
..#
#..

Sample Output 1

10

Some of the pairs satisfying the condition are ((1, 2), (3, 3)) and ((3, 1), (3, 2)), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left.

Sample Input 2

2 4
....
....

Sample Output 2

0

Sample Input 3

4 3
###
###
...
###

Sample Output 3

6

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

    #include <stdio.h>
    #include <memory.h>
    long paint(char* field, int x, int y)
    {
      int stack[160000];
      long odd = 0;
      long even = 0;
      int p = y * 402 + x;
      if((x + y) % 2)
      {
        odd = 1;
      }
      else
      {
        even = 1;
      }
      int color = field[p];
      int stn = 1;
      stack[0] = p;
      field[p] = 0;
      while(stn)
      {
        stn--;
        int pos = stack[stn];
        int way[4] = {-402, -1, 1, 402};
        int w;
        for(w = 0 ; w < 4 ; w++)
        {
        	int newpos = pos + way[w];
    	    if(field[newpos] == color)
    	    {
    	      field[newpos] = 0;
    	      if((newpos / 402 + newpos % 402) % 2)
    	      {
    	        odd++;
    	      }
    	      else
    	      {
    	        even++;
    	      }
    	      stack[stn] = newpos;
    	      stn++;
    	    }
        }
      }
      return odd * even;
    }
     
    int main()
    {
      char field[402][402];
      int h, w;
      int x, y;
      scanf("%d%d", &h, &w);
      memset(field, 0, 402*402);
      for(y = 1 ; y <= h ; y++)
      {
        scanf("%s", field[y]+1);
        for(x = 1 ; x <= w ; x++)
        {
          if((x + y) % 2 == 0)
          {
            field[y][x] = field[y][x] ^ ('#' ^ '.');
          }
        }
      }
      long sum = 0;
      for(y = 1 ; y <= h ; y++)
      {
        for(x = 1 ; x <= w ; x++)
        {
        	if(field[y][x])
        	{
        		sum += paint((char*)field, x, y);
        	}
        }
      }
      printf("%ld\n", sum);
      return 0;
    }