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.

Examples

Input

3 3
.#.
..#
#..


Output

10


Input

3 3
.#.
..#
..


Output

10


Input

2 4
....
....


Output

0


Input

4 3


...


Output

6
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

	int H, W;
	std::vector<std::string> S;

void count_area(long* n, int x, int y, char ch)
{
	S[y][x] = ' ';
	++n[(x+y)%2];
	if (x > 0 && S[y][x-1] == ch) {
		count_area(n, x-1, y, ch);
	}
	if (x < W-1 && S[y][x+1] == ch) {
		count_area(n, x+1, y, ch);
	}
	if (y > 0 && S[y-1][x] == ch) {
		count_area(n, x, y-1, ch);
	}
	if (y < H-1 && S[y+1][x] == ch) {
		count_area(n, x, y+1, ch);
	}
}

int main()
{

	std::cin >> H >> W;
	S.reserve(H);
	for (int i = 0; i < H; ++i) {
		std::string s;
		std::cin >> s;
		S.push_back(s);
	}

	for (int i = 0; i < H; ++i) {
		auto& s = S[i];
		for (int j = i%2; j < s.size(); j += 2) {
			if (s[j] == '.')
				s[j] = '#';
			else
				s[j] = '.';
		}
	}

	long total = 0;
	for (int y = 0; y < H; ++y) {
		for (int x = 0; x < W; ++x) {
			if (S[y][x] != ' ') {
				long n[2] = {0,0};
				count_area(n, x, y, S[y][x]);
				total += n[0] * n[1];
			}
		}
	}
	std::cout << total << "\n";
}
