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 IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define int long long

const int N=405;

int n, m, ct=0, ans=0, bb=0, w=0;
int vis[N][N];
char a[N][N];
int b[N][N];

void dfs(int x, int y, int c)
{
	if(x<1 || y<1 || x>n || y>m)
		return;
	if(vis[x][y])
		return;
	if(b[x][y]!=c)
		return;
	bb+=(b[x][y]);
	w+=(!b[x][y]);
	vis[x][y]=1;
	ct++;
	dfs(x+1, y, c^1);
	dfs(x-1, y, c^1);
	dfs(x, y+1, c^1);
	dfs(x, y-1, c^1);
}

int32_t main()
{
	IOS;
	cin>>n>>m;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			cin>>a[i][j];
			b[i][j]=(a[i][j]=='#');
		}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			if(vis[i][j])
				continue;
			ct=0;
			bb=0;
			w=0;
			dfs(i, j, b[i][j]);
			ans+=bb*w;
		}
	}
	cout<<ans;
	return 0;
}