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>
using namespace std;

int h, w, dx[4] = { 0, 1, 0, -1 }, dy[4] = { 1, 0, -1, 0 };
long long c1, c2;
string s[409];
bool vs[409][409];

void dfs(int x, int y)
{
	vs[y][x] = true;
	if (s[y][x] == '#') c1++;
	else c2++;
	for (int i = 0; i < 4; i++) {
		int tx = x + dx[i], ty = y + dy[i];
		if (tx < 0 || tx >= w || ty < 0 || ty >= h || s[y][x] == s[ty][tx] || vs[ty][tx]) continue;
		dfs(tx, ty);
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	cin >> h >> w;
	for (int i = 0; i < h; i++)
		cin >> s[i];
	long long ans = 0;
	for (int i = 0; i < h; i++)
		for (int j = 0; j < w; j++) {
			if (vs[i][j]) continue;
			c1 = c2 = 0;
			dfs(j, i);
			ans += c1 * c2;
		}
	cout << ans << '\n';
	return 0;
}