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 <algorithm>
#include <vector>
#include <string>
using namespace std;
#define int long long
char c[500][500];
bool b[500][500];
typedef pair<int, int>p;
p f1(p a, p b) { return make_pair(a.first + b.first, a.second + b.second); }
p f(int i, int j) {
	p r = make_pair(0, 0);
	if (b[i][j])return r;
	b[i][j] = 1;
	char ch = (c[i][j] == '.' ? '#' : '.');
	if (c[i][j] == '.')r.first++; else r.second++;
	if (c[i - 1][j] == ch && !b[i - 1][j])r = f1(r, f(i - 1, j));
	if (c[i + 1][j] == ch && !b[i + 1][j])r = f1(r, f(i + 1, j));
	if (c[i][j - 1] == ch && !b[i][j - 1])r = f1(r, f(i, j - 1));
	if (c[i][j + 1] == ch && !b[i][j + 1])r = f1(r, f(i, j + 1));
	return r;
}
signed main() {
	int h, w; cin >> h >> w;
	for (int i = 1; i <= h; i++) {
		for (int j = 1; j <= w; j++) {
			cin >> c[i][j];
		}
	}
	int ans = 0;
	for (int i = 1; i <= h; i++) {
		for (int j = 1; j <= w; j++) {
			if (!b[i][j]) {
				p p1 = f(i, j);
				ans += p1.first * p1.second;
			}
		}
	}
	cout << ans << endl;
	cin >> c[0][0];
}