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>
using namespace std;
int par[160010];
int a[160010]={0};
int b[160010]={0};
int n,m;
char c[410][410];
int d1[4]={0,0,1,-1};
int d2[4]={1,-1,0,0};
inline int pare(int x)
{
	if (par[x]==x) return x;
	else return pare(par[x]);
}
void join(int x,int y)
{
	x=pare(x),y=pare(y);
	if (x==y) return;
	par[x]=y;
	a[y]+=a[x];
	b[y]+=b[x];
}
int main()
{
	cin >>n>>m;
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<m;j++)
		{
			cin >>c[i][j];
			(c[i][j]=='.' ? a[i*m+j] : b[i*m+j])=1;
			par[i*m+j]=i*m+j;
		}
	}
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<m;j++)
		{
	        int t=i*m+j;
	        for (int k=0;k<4;k++)
	        {
	        	int xi=i+d1[k];
	        	int yi=j+d2[k];
	        	if (xi>=0 && yi>=0 && xi<n && yi<m && c[xi][yi]!=c[i][j])
	        	    join(t,xi*m+yi);
			}
		}
	}
	long long res=0;
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<m;j++)
		{
	        int t=i*m+j;
	        if (par[t]!=t) continue;
	        res+=(long long)a[t]*b[t];
		}
	}
	cout <<res;
	return 0;
}