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 <bits/stdc++.h>
using namespace std;
#define int long long
char arr[405][405];
bool vis[405][405];
int h, w;
int cb;
int cw;

bool f(int i, int j, int l, int m) {
	if (l < 0 || l >= h || m < 0 || m >= w) {
		return 0;
	}
	if (vis[l][m] == 1) {
		return 0;
	}
	if (arr[i][j] == arr[l][m]) {
		return 0;
	}
	return 1;
}

void dfs(int i, int j) {
	vis[i][j] = 1;
	// cout << i << " " << j << endl;
	if (arr[i][j] == '#') {
		cb++;
	}
	else {
		cw++;
	}
	if (f(i, j, i + 1, j)) {
		dfs(i + 1, j);
	}
	if (f(i, j, i, j + 1)) {
		dfs(i, j + 1);
	}
	if (f(i, j, i - 1, j)) {
		dfs(i - 1, j);
	}
	if (f(i, j, i, j - 1)) {
		dfs(i, j - 1);
	}
}


int32_t main() {
	cin >> h >> w;
	for (int i = 0; i < h; i++) {
		for (int j = 0; j < w; j++) {
			cin >> arr[i][j];
		}
	}
	int ans = 0;
	for (int i = 0; i < h; i++) {
		for (int j = 0; j < w; j++) {
			if (vis[i][j] == 0) {
				dfs(i, j);
				// cout << "cb=" << cb << " " << "cw=" << cw << endl;
				ans += (cb * cw);
				cb = 0;
				cw = 0;
			}
		}
	}
	cout << ans;
	return 0;
}