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<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
int n,m,pa[200005],num1[200005],num2[200005];
bool a[405][405];
ll ans;
inline bool read()
{
	char c=getchar();
	while(c!='#'&&c!='.')c=getchar();
	return c=='#';
}
int find(int x)
{
	return x==pa[x]?x:pa[x]=find(pa[x]);
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++)
	for(int j=0;j<m;j++)
	{
		a[i][j]=read();
		if(a[i][j])num1[i*m+j+1]++;
		else num2[i*m+j+1]++;
		pa[i*m+j+1]=i*m+j+1;
	}
	for(int i=0;i<n;i++)
	for(int j=0;j<m;j++)
	{
		if(i!=n-1&&(a[i][j]^a[i+1][j]))
		{
			int u=find(i*m+j+1),v=find((i+1)*m+j+1);
			if(u!=v)
			{
				num1[u]+=num1[v];
				num2[u]+=num2[v];
				pa[v]=u;
			}
		}
		if(j!=m-1&&(a[i][j]^a[i][j+1]))
		{
			int u=find(i*m+j+1),v=find(i*m+j+1+1);
			if(u!=v)
			{
				num1[u]+=num1[v];
				num2[u]+=num2[v];
				pa[v]=u;
			}
		}
	}
	for(int i=1;i<=n*m;i++)
	  if(find(i)==i)
	    ans+=1ll*num1[i]*num2[i];
	printf("%lld\n",ans);
	return 0;
}