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 <vector>
#include <algorithm>
using namespace std;
typedef long long int lli;
int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
bool used[400][400] = {};
lli h,w;
void dfs(int y, int x, char bw, lli &bl, lli &wh, vector<string> &field){
	if(used[y][x]) return;
	used[y][x] = true;
	if(field[y][x] == '.') wh++;
	else bl++;
	for(int d=0; d<4; d++){
		int ny = y+dy[d];
		int nx = x+dx[d];
		if(ny<0 || ny>=h || nx<0 || nx>=w) continue;
		if(field[y][x] == field[ny][nx]) continue;
		dfs(ny, nx, field[y][x], bl, wh, field);
	}
}

int main(){
	cin >> h >> w;
	vector<string> field(h);
	for(int i=0; i<h; i++){
		cin >> field[i];
	}
	lli ans = 0;
	for(int i=0; i<h; i++){
		for(int j=0; j<w; j++){
			lli bl=0, wh=0;
			dfs(i, j, 0, bl, wh, field);
			ans += bl*wh;
		}
	}
	cout << ans << endl;
	return 0;
}