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 <cstdio>
#include <vector>
#define N 405
using namespace std;

typedef long long ll;
ll n, m, num, cnt[160005], s[160005], z[N][N];
ll dr[4] = {-1, 0, 0, 1}, dc[4] = {0, -1, 1, 0};
bool v[N][N];
char a[N][N];
//vector<ll> x, y;
void f(ll p, ll q) {
	ll i, j, x, y;
	if (p < 1 || p > n || q < 1 || q > m || v[p][q] || a[p][q] != '#') return;
	v[p][q] = 1;
	s[num]++;
	for (i = 0; i < 4; i++) {
		x = p + dr[i];
		y = q + dc[i];
		if (a[x][y] != '.') continue;
		if (z[x][y] != num) {
			cnt[num]++;
			z[x][y] = num;
			for (j = 0; j < 4; j++) {
				f(x + dr[j], y + dc[j]);
			}
		}
	}
}
int main()
{
	ll i, j;
	cin >> n >> m;
	for (i = 1; i <= n; i++) {
		scanf ("%s", &a[i][1]);
	}
	for (i = 1; i <= n; i++) {
		for (j = 1; j <= m; j++) {
			if (v[i][j] || a[i][j] != '#') continue;
			num++;
			f(i, j);
		}
	}
	ll ans = 0;
	for (i = 1; i <= num; i++) {
		ans += s[i] * cnt[i];
	}
	cout << ans;
    return 0;
}
