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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <math.h>

int h,w,visited[500][500],q,k;
long long ans;
char s[500][500];


int dfsw(int a,int b);
int dfsb(int a,int b);

int main(void){
    scanf("%d%d",&h,&w);
    for (int i=0; i<h; i++){
        scanf("%s",s[i]);
    }
    for (int i=0; i<h; i++){
        for (int j=0; j<w; j++){
            q=0;
            k=0;
            if (s[i][j]=='#' && visited[i][j]==0){
                q++;
                visited[i][j]=1;
                dfsb(i,j);
                
                
            }
            ans+=q*k;
        }
    }
    printf("%lld\n",ans);
    return 0;
}

int dfsb(int a,int b){
    if (a+1<h  && s[a+1][b]=='.' && visited[a+1][b]==0){
        k++;
        visited[a+1][b]=1;
        dfsw(a+1,b);
    }
    if (b+1<w && s[a][b+1]=='.' && visited[a][b+1]==0){
        k++;
        visited[a][b+1]=1;
        dfsw(a,b+1);
    }
    if (a-1>=0 && s[a-1][b]=='.' && visited[a-1][b]==0){
        k++;
        visited[a-1][b]=1;
        dfsw(a-1,b);
    }
    if (b-1>=0 && s[a][b-1]=='.' && visited[a][b-1]==0){
        k++;
        visited[a][b-1]=1;
        dfsw(a,b-1);
    }
    return 0;
}

int dfsw(int a,int b){
    if (a+1<h && s[a+1][b]=='#' && visited[a+1][b]==0){
        q++;
        visited[a+1][b]=1;
        dfsb(a+1,b);
    }
    if (b+1<w && s[a][b+1]=='#' && visited[a][b+1]==0){
        q++;
        visited[a][b+1]=1;
        dfsb(a,b+1);
    }
    if (a-1>=0 && s[a-1][b]=='#' && visited[a-1][b]==0){
        q++;
        visited[a-1][b]=1;
        dfsb(a-1,b);
    }
    if (b-1>=0 && s[a][b-1]=='#' && visited[a][b-1]==0){
        q++;
        visited[a][b-1]=1;
        dfsb(a,b-1);
    }
    return 0;
}