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 ll long long

ll ans = 0;
int m[500][500];
char s[500][500];
ll bk = 0;
ll wh = 0;
int h, w;

void dfs(int x, int y, char be){
	if(x < 0 || y < 0 || x >= h || y >= w) return;
	if(m[x][y] == 1) return;
	if(be != s[x][y]){
		if(s[x][y] == '#'){
			bk++;
		}else{
			wh++;
		}
	}else{
		return;
	}
	m[x][y] = 1;
	//cout << x << y << " " << bk << wh << endl;
	dfs(x + 1, y, s[x][y]);
	dfs(x, y + 1, s[x][y]);
	dfs(x - 1, y, s[x][y]);
	dfs(x, y - 1, s[x][y]);
}

int main(void){

	cin >> h >> w;
	for(int i = 0; i < h; ++i){
		for(int j = 0; j < w; ++j){
			cin >> s[i][j];
		}
	}
	for(int i = 0; i < h; ++i){
		for(int j = 0; j < w; ++j){
			dfs(i, j, 'x');
			ans += bk * wh;
			//cout << bk << wh << endl;
			bk = 0;
			wh = 0;
		}
	}
	cout << ans << endl;
	return 0;
}