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>
using namespace std;
#define ll long long
#define pb push_back
#define mp make_pair
const int M=405;
const int N=M*M;
vector<int> E[N];
bool was[N];
int cnt[2],col[N];
void DFS(int u)
{
	was[u]=1;
	cnt[col[u]]++;
	for(int v:E[u]) if(!was[v]) DFS(v);
}
void AddEdge(int u, int v)
{
	E[u].pb(v);
	E[v].pb(u);
}
char Base[M][M];
int Move[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int main()
{
	int n,m,k;
	scanf("%i %i",&n,&m);
	k=n*m;
	for(int i=1;i<=n;i++)
	{
		scanf("%s",Base[i]+1);
		for(int j=1;j<=m;j++)
		{
			col[(i-1)*m+j]=Base[i][j]=='.'?0:1;
			for(int mo=0;mo<4;mo++)
			{
				int x=i+Move[mo][0];
				int y=j+Move[mo][1];
				if((Base[i][j]=='#' && Base[x][y]=='.') || (Base[x][y]=='#' && Base[i][j]=='.'))
				{
					AddEdge((i-1)*m+j,(x-1)*m+y);
				}
			}
		}
	}
	ll ans=0;
	for(int i=1;i<=k;i++) if(!was[i])
	{
		cnt[0]=cnt[1]=0;
		DFS(i);
		ans+=(ll)cnt[0]*cnt[1];
	}
	printf("%lld\n",ans);
	return 0;
}
