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 <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>

int n, m;
long long ans;
char map[401][401];
bool vis[401][401];
int dx[4] = {0, 0, -1, 1}, dy[4] = {1, -1, 0, 0};
int sdx[200001], sdy[200001];

void bfs(int sx, int sy) {
	int head = 0, tail = 1, w = 0, h = 0;
	sdx[tail] = sx, sdy[tail] = sy;
	if (map[sx][sy] == '#') ++w;
	else ++h;
	while(head < tail) {
		++head;
		for (int i = 0; i <= 3; ++i) {
			int c = sdx[head] + dx[i], d = sdy[head] + dy[i];
			if (c > 0 && c <= n && d > 0 && d <= m && !vis[c][d] && map[sdx[head]][sdy[head]] != map[c][d]) {
				if (map[c][d] == '#') ++w;
				else ++h;
				vis[c][d] = 1, ++tail;
				sdx[tail] = c, sdy[tail] = d;
			}
		}
	}
	ans += 1ll * w * h;
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			std::cin >> map[i][j];
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			if (!vis[i][j]) vis[i][j] = 1, bfs(i, j);
		}
	}
	printf("%lld\n", ans);
	return 0;
}