Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>

using namespace std;

int h,w;
char s[405][405];
bool v[405][405];
long long a,b,ans;
void dfs(int x,int y){
	v[x][y]=1;
	if(s[x][y]=='#')a++;
	else b++;
	if(x>1&&v[x-1][y]==0&&s[x-1][y]!=s[x][y])dfs(x-1,y);
	if(x<h&&v[x+1][y]==0&&s[x+1][y]!=s[x][y])dfs(x+1,y);
	if(y>1&&v[x][y-1]==0&&s[x][y-1]!=s[x][y])dfs(x,y-1);
	if(y<w&&v[x][y+1]==0&&s[x][y+1]!=s[x][y])dfs(x,y+1);
}
int main()
{
	cin>>h>>w;
	for(int i=1;i<=h;i++)
		scanf("%s",s[i]+1);
	for(int i=1;i<=h;i++)
		for(int j=1;j<=w;j++)
			if(v[i][j]==0){
				a=b=0;
				dfs(i,j);
				ans+=a*b;
			}
	printf("%lld",ans);
}