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 <stdio.h>
int h, w;
char s[405];
int a[402][402];
int b[402][402];
long long br, wh;
long long ans;
void dfs(int x, int y) {
	b[x][y] = 1;
	if (a[x][y])br++;
	else wh++;
	if (!b[x][y + 1] && a[x][y] + a[x][y + 1] == 1) {
		dfs(x, y + 1);
	}
	if (!b[x][y - 1] && a[x][y] + a[x][y - 1] == 1) {
		dfs(x, y - 1);
	}
	if (!b[x + 1][y] && a[x][y] + a[x + 1][y] == 1) {
		dfs(x + 1, y);
	}
	if (!b[x - 1][y] && a[x][y] + a[x - 1][y] == 1) {
		dfs(x - 1, y);
	}
	return;
}
int main(){
	scanf("%d%d", &h, &w);
	for (int i = 0; i < 402; i++) {
		for (int j = 0; j < 402; j++) {
			a[i][j] = 2;
		}
	}
	for (int i = 0; i < h; i++) {
		scanf("%s", s);
		for (int j = 0; j < w; j++) {
			if (s[j] == '.')a[i + 1][j + 1] = 0;
			if (s[j] == '#')a[i + 1][j + 1] = 1;
		}
	}
	for (int i = 1; i <= h; i++) {
		for (int j = 1; j <= w; j++) {
			br = 0; wh = 0;
			dfs(i, j);
			ans += br * wh;
		}
	}
	printf("%lld\n", ans);
	return 0;
}