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>
#define ll long long
using namespace std;
int n,m;
char c[410][410];
bool visit[410][410]={0};
ll res=0;
int t,tt;
void dfs(int x,int y)
{
	(c[x][y]=='.' ? t : tt)++;
	visit[x][y]=1;
	if (x>0 && c[x-1][y]!=c[x][y] && !visit[x-1][y]) dfs(x-1,y);
	if (y>0 && c[x][y-1]!=c[x][y] && !visit[x][y-1]) dfs(x,y-1);
	if (x<n-1 && c[x+1][y]!=c[x][y] && !visit[x+1][y]) dfs(x+1,y);
	if (y<m-1 && c[x][y+1]!=c[x][y] && !visit[x][y+1]) dfs(x,y+1);
}
int main()
{
	cin >>n>>m;
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<m;j++)
		{
			cin >>c[i][j];
		}
	}
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<m;j++)
		{
			if (!visit[i][j])
			{
				t=tt=0;
				dfs(i,j);
				res+=(ll)t*tt;
			}
		}
	}
	cout <<res;
	return 0;
}