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;
typedef long long ll;
const int M=2e6+4;
char a[M];
int cnt=1,vis[M];
ll x,y,ans=0;
vector<int>v[M];
void dfs(int u)
{
	if(vis[u]) return;
	vis[u]=1;
	if(a[u]=='.') x++;
	else if(a[u]=='#') y++;
	for(auto x:v[u])	dfs(x);
}
int main()
{
	int n,m;
	cin>>n>>m;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			char ch;
			cin>>ch;
			a[cnt]=ch;
			if(cnt%m!=1 || m==1)
			{
				if(a[cnt-1]!=ch)
				{
					v[cnt].push_back(cnt-1);
					v[cnt-1].push_back(cnt);
				}
			}
			if(cnt>m)
			{
				if(a[cnt-m]!=ch)
				{
					v[cnt].push_back(cnt-m);
					v[cnt-m].push_back(cnt);
				}
			}
			cnt++;
		}
	}
	for(int i=1;i<=n*m;i++)
	{
		if(!vis[i])
		{
			x=y=0;
			dfs(i);
			//cout<<i<<" "<<x<<" "<<y<<endl;
			ans+=x*y;
		}
	}
	cout<<ans;
	return 0;
}