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 maxn 500
#define ll long long
char str[maxn][maxn];
int n,m,i,j,s1,s2;
int mv[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
bool mark[maxn][maxn];
ll ans=0;
using namespace std;
void dfs(int x,int y)
{
	int i,px,py;
	mark[x][y]=false;
	if (str[x][y]=='.') s2++;
		else s1++;
	for (i=0;i<4;i++)
	{
		px=x+mv[i][0];
		py=y+mv[i][1];
		if (px && py && (px<=n) && (py<=m) && mark[px][py] && (str[x][y]!=str[px][py])) dfs(px,py);
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for (i=1;i<=n;i++)
		scanf("%s",str[i]+1);
	memset(mark,true,sizeof(mark));
	for (i=1;i<=n;i++)
		for (j=1;j<=m;j++)
			if (mark[i][j])
			{
				s1=s2=0;
				dfs(i,j);
				ans+=1LL*s1*s2;
			}
	printf("%lld",ans);
	return 0;
}