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;
#define REP(i,n) for(int i=0; i<(n); i++)
#define REP2(i,x,n) for(int i=x; i<(n); i++)
#define ALL(n) begin(n),end(n)
struct cww{cww(){ios::sync_with_stdio(false);cin.tie(0);}}star;
const long long INF = numeric_limits<long long>::max();
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int main() {
  int H, W;
  cin >> H >> W;
  string S[H];
  REP(i, H) cin >> S[i];

  bool seen[H][W];
  REP(i, H) REP(j, W) seen[i][j] = false;
  queue<pair<int, int>> todo;

  int dx[4] = {1,0,-1,0};
  int dy[4] = {0,1,0,-1};
  long long ans = 0;
  REP(i, H) REP(j, W) if (seen[i][j] == false) {
    // cout << i << ' ' << j << endl;
    long long cnt[] = {0, 0};
    seen[i][j] = true;
    todo.push(make_pair(i, j));
    while (todo.size() > 0) {
      pair<int, int> here = todo.front(); todo.pop();
      // cout << here.first << endl;
      if (S[here.first][here.second] == '#') cnt[0]++;
      else cnt[1]++;
      REP(k, 4) {
        if (0 <= here.first+dy[k] && here.first+dy[k] < H && 0 <= here.second+dx[k] && here.second+dx[k] < W && S[here.first][here.second] != S[here.first+dy[k]][here.second+dx[k]] && seen[here.first+dy[k]][here.second+dx[k]] == false) {
          seen[here.first+dy[k]][here.second+dx[k]] = true;
          todo.push(make_pair(here.first+dy[k], here.second+dx[k]));
        }
      }
    }
  ans += cnt[0] * cnt[1];
  }
  cout << ans << endl;


  return 0;
}