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;
typedef pair<int,int> pii;

string grid[405];
int H,W;
int xi[] = {1,0,-1,0};
int yi[] = {0,1,0,-1};
bool visited[405][405];
pii solve(int x, int y){
	visited[x][y] = true;
	int d=0,h=0;
	if(grid[x][y]=='.'){
		d++;
		for(int i=0;i<4;i++){
			int nx = x+xi[i], ny = y+yi[i];
			if(nx>=0 and ny>=0 and nx<H and ny<W and !visited[nx][ny] and grid[nx][ny]=='#'){
				pii p = solve(nx,ny);
				d+=p.first;
				h+=p.second;
			}
		}
	}else{
		h++;
		for(int i=0;i<4;i++){
			int nx = x+xi[i], ny = y+yi[i];
			if(nx>=0 and ny>=0 and nx<H and ny<W and !visited[nx][ny] and grid[nx][ny]=='.'){
				pii p = solve(nx,ny);
				d+=p.first;
				h+=p.second;			
			}
		}

	}
	return pii(d,h);
}

int main(){
	long long ans=0;
	cin >> H >> W;
	for(int i=0;i<H;i++)
		cin >> grid[i];
	for(int i=0;i<H;i++){
		for(int j=0;j<W;j++){
			if(!visited[i][j]){
				pii p = solve(i,j);
				ans += (long long)p.first * p.second;
			}
		}
	}
	cout << ans;

}