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 <iostream>
#include <vector>
using namespace std;

#define rep(i, n) for (int i = 0; i < n; i++)

struct UF {
	vector<int> black, white, par;
	UF(int sz) : black(sz), white(sz), par(sz) {
		rep(i, sz) par[i] = i;
	}
	int parent(int i) {
		if (par[i] == i) return i;
		return par[i] = parent(par[i]);
	}
	void uni(int i, int j) {
		int pi = parent(i), pj = parent(j);
		if (pi == pj) return;
		black[pj] += black[pi];
		white[pj] += white[pi];
		par[pi] = pj;
	}
};

int main() {
	int H, W; cin >> H >> W;
	vector<vector<char>> S(H);
	UF uf = UF(H * W);
	auto idx = [W](int i, int j) {
		return i * W + j;
	};
	rep(i, H) {
		S[i].resize(W);
		rep(j, W) {
			cin >> S[i][j];
			if (S[i][j] == '#') uf.black[idx(i, j)] = 1;
			else uf.white[idx(i, j)] = 1;
		}
	}
	rep (i, H) {
		rep (j, W) {
			char c = S[i][j];
			int index = idx(i, j);
			if (0 <= i - 1 && c != S[i - 1][j]) uf.uni(index, idx(i - 1, j));
			if (0 <= j - 1 && c != S[i][j - 1]) uf.uni(index, idx(i, j - 1));
		}
	}
	uint64_t ans = 0;
	rep (i, H * W) if (uf.par[i] == i) ans += (uint64_t)uf.black[i] * uf.white[i];
	cout << ans << endl;
}