Problem: 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

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstdlib>
#include<deque>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int>P;
const int len=1e3+5;
const ll mod=998244353;
const ull seed=131;
int dx[4]={1,-1,0,0};
int dy[4]={0,0,-1,1};
char dd[len][len];
int n,m;
int x,y;
void dfs(int i,int j)
{
	if(dd[i][j]=='.')y++;
	else if(dd[i][j]=='#')x++;
	char ch=dd[i][j];
	dd[i][j]='!';
	for(int u=0;u<4;++u)
	{
		int xx=i+dx[u];
		int yy=j+dy[u];
		if(dd[xx][yy]!='!'&&xx>=1&&xx<=n&&yy>=1&&yy<=m&&ch!=dd[xx][yy])
		{
			dfs(xx,yy);
		}
	}
}
int main()
{	
	cin>>n>>m;
	for(int i=1;i<=n;++i)scanf("%s",dd[i]+1);
	ll ans=0;
	for(int i=1;i<=n;++i)
	{
		for(int j=1;j<=m;++j)
		{
			if(dd[i][j]!='!')
			{
				x=0;
				y=0;
				dfs(i,j);
				ans+=(ll)x*y;
			}
		}
	}
	cout<<ans<<endl;
}