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 <bits/stdc++.h>
using namespace std;
using ll=long long;
using P = pair<ll,ll>;
ll H,W;

bool onBoard(int y,int x){
    if(y>=0 && y<H && x>= 0 && x<W){
        return true;
    }
    return false;
}


int main() {
    cin >> H >> W;
    vector<string> S(H);
    for(int i=0;i<H;i++){
        cin >> S[i];
    }
    bool alreadyChecked[H][W];
    for(int i=0;i<H;i++)for(int j=0;j<W;j++){
        alreadyChecked[i][j] = false;
    }
    int dy[] = {1,0,-1,0};
    int dx[] = {0,1,0,-1};
    ll ans =0;
    for(int sy =0;sy < H;sy++)for(int sx = 0;sx<W;sx++){
        if(S[sy][sx] == '.' || alreadyChecked[sy][sx]){
            continue;
        }
        // cerr <<"sy=" << sy <<"sx="<<sx << endl;
        ll sameNum =1;
        ll whiteNum = 0;
        queue<P> q;
        q.push(P(sy,sx));
        alreadyChecked[sy][sx] = true;
        while(!q.empty()){
            P p = q.front();q.pop();
            for(int i=0;i<4;i++){
                ll ny = p.first + dy[i];
                ll nx = p.second + dx[i];
                // cerr <<"ny=" <<ny <<"nx=" << nx << endl;
                if(!onBoard(ny,nx) || alreadyChecked[ny][nx] || (S[p.first][p.second] == S[ny][nx])){
                    continue;
                }
                
                if(S[ny][nx] == '#'){
                    sameNum++;
                }else{
                    whiteNum++;
                }
                alreadyChecked[ny][nx] = true;
                q.push(P(ny,nx));
            }
        }
        // cerr <<"sameNum=" << sameNum <<"whiteNum=" << whiteNum << endl;
        ans += (sameNum * whiteNum);
    }
    cout << ans << endl;
    return 0;
}