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<bits/stdc++.h>
#define pii pair<int,int>
#define mp make_pair
#define p push
#define f first
#define s second
using namespace std;
char b[444][444];
bool vis[444][444];
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1},n,m;
long long ans;
int main()
{
	cin>>n>>m;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			cin>>b[i][j];
		}
	}
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			if(vis[i][j])
			{
				continue;
			}
			queue<pii>q;
			q.p(mp(i,j));
			vis[i][j]=true;
			long long cnt1=0,cnt2=0;
			while(!q.empty())
			{
				pii i=q.front();
				q.pop();
				if(b[i.f][i.s]=='.')
				{
					cnt1++;
				}
				else
				{
					cnt2++;
				}
				for(int d=0;d<4;d++)
				{
					int x=i.f+dx[d],y=i.s+dy[d];
					if(0<=x && x<n && 0<=y && y<m)
					{
						if(b[i.f][i.s]!=b[x][y] && !vis[x][y])
						{
							vis[x][y]=true;
							q.p(mp(x,y));
						}
					}
				}
			}
			ans+=cnt1*cnt2;
		}
	}
	cout<<ans<<endl; 
	return 0;
}