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>
#include <algorithm>
using namespace std;
using ll = long long;

int h, w;
char s[400][401];
int dx[] = { -1, 1, 0, 0 }, dy[] = { 0, 0, -1, 1 };
void dfs(int x, int y, ll& black, ll& white) {
	(s[x][y] == '#' ? black : white)++;
	char cur = s[x][y];
	s[x][y] = 0;
	for (int i = 0; i < 4; i++) {
		int nx = x + dx[i], ny = y + dy[i];
		if (0 <= nx && nx < h && 0 <= ny && ny < w &&
			s[nx][ny] != 0 && cur != s[nx][ny])
			dfs(nx, ny, black, white);
	}
}
int main() {
	scanf("%d %d", &h, &w);
	for (int i = 0; i < h; i++)
		scanf("%s", s[i]);
	ll total = 0;
	for (int i = 0; i < h; i++)
		for (int j = 0; j < w; j++) {
			ll b = 0, w = 0;
			if (s[i][j] != 0) {
				dfs(i, j, b, w);
				total += b * w;
			}
		}
	printf("%lld\n", total);
}