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: Time Limit Exceeded)

#include<bits/stdc++.h>
#define rep(i,N) for(ll (i)=0;(i)<(N);(i)++)
#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const int mod = 1000000007;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};

int main() {
  int h, w;
  cin >> h >> w;
  vector<string> s(h);
  rep(i, h) cin >> s[i];
  int ans = 0;
  rep(x, h) rep(y, w) {
    if(s[x][y] != '.') continue;
    queue<P> q;
    vector<vector<bool>> used(h, vector<bool>(w));
    int cnt = 0;
    auto push = [&] (int i, int j, char c) {
      if(used[i][j]) return;
      if(s[i][j] == c) return;
      used[i][j] = true;
      if(s[i][j] == '#') {
        ++cnt;
      }
      q.push(P(i, j));
    };
  
    push(x, y, '#');
    while(!q.empty()) {
      auto v = q.front(); q.pop();
      int i = v.first, j = v.second;
      rep(d, 4) {
        int nx = dx[d] + i;
        int ny = dy[d] + j;
        char nc = s[i][j];
        if(0 <= nx && nx < h && 0 <= ny && ny < w) push(nx, ny, nc);
      }
    }
    ans += cnt;
  }
  cout << ans << endl;
}