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:
#pragma GCC optimize ("Ofast","unroll-loops")
#include <bits/stdc++.h>
#define pb push_back
#define eb emplace_back
#define MP make_pair
#define F first
#define S second
#define MEM(a,b) memset(a,b,sizeof a)
#define Tie ios::sync_with_stdio(0),cin.tie(0);
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

ll h,w,dx[4]={0,1,0,-1},dy[4]={1,0,-1,0},l,r,ans;
string s[400];
bitset<400> u[400];

inline bool c(int x,int y)
{
	if(x<0||y<0||x>=h||y>=w||u[x][y])
		return 0;
	return 1;
}

void dfs(int x,int y)
{
	u[x][y]=1;
	if(s[x][y]=='.')
		r++;
	else
		l++;
	for(int i=0;i<4;i++)
		if(c(x+dx[i],y+dy[i])&&((s[x][y]=='#'&&s[x+dx[i]][y+dy[i]]=='.')||(s[x][y]=='.'&&s[x+dx[i]][y+dy[i]]=='#')))
			dfs(x+dx[i],y+dy[i]);
}

int main()
{
	cin>>h>>w;
	for(int i=0;i<h;i++)
		cin>>s[i];
	for(int i=0;i<h;i++)
		for(int j=0;j<w;j++)
			if(!u[i][j]&&s[i][j]=='#')
				l=r=0,dfs(i,j),ans+=l*r;
	cout << ans<<'\n';
	
}