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>
#define ll long long int
using namespace std;
int x[4]={0,0,-1,1};
int y[4]={1,-1,0,0};
ll r,c;
string s[405];
ll vis[405][405];
bool chk(ll s1,ll s2)
{
	return s1>=0 && s1<r && s2>=0 && s2<c;
}
ll ans,white,black;
void dfs(ll s1,ll s2, char prev){
	s[s1][s2]='o';
	char next;
	if(prev=='#')
	{
		black++;
		next='.';
	}
	else{
		white++;
		next='#';
	}
	for(ll i=0;i<4;i++)
	{
			ll pntx=s1+x[i];
			ll pnty=s2+y[i];
			if(chk(pntx,pnty) && s[pntx][pnty]==next){
				dfs(pntx,pnty,next);
			}
		}
}
int main()
{
	ll i,j;
	cin>>r>>c;
	for(i=0;i<r;i++)
	{
		cin>>s[i];
	}
	for(i=0;i<r;i++)
	{
		for(j=0;j<c;j++)
		{
			if(s[i][j]=='#')
			{
				white=0;
				black=0;
				dfs(i,j,'#');
				ans+=white*black;
				//memset(vis,0,sizeof vis);
			}
		}
	}
	cout<<ans;
 	return 0;
}