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>
using namespace std;
typedef long long ll;

ll movx[] = {0,1,0,-1},movy[] = {1,0,-1,0}; 

int main()
{
	ll h,w;
	cin >> h >> w;
	vector<string> s(h);
	for(ll i = 0;i<h;i++)
	{
		cin >> s[i];
	}
	vector<vector<ll>> COL(h,vector<ll>(w));
	queue<pair<ll,ll>> Q;
	ll nowC = 1;
	ll cntB = 0,cntW = 0;
	ll ans = 0;
	for(ll i = 0;i<h;i++)
	{
		for(ll j = 0;j<w;j++)
		{
			if(COL[i][j]==0&&s[i][j]=='#')
			{
				COL[i][j] = 1;
				Q.push(make_pair(i,j));
				cntB =1;
				cntW = 0;
				while(!Q.empty())
				{
					pair<ll,ll> now = Q.front();
					Q.pop();
					for(ll ii = 0;ii< 4;++ii)
					{
						ll nx = now.first+movx[ii];
						ll ny = now.second+movy[ii];
						if(0<=nx&&nx<h&&0<=ny&&ny<w)
						{
							if(s[now.first][now.second]!=s[nx][ny]&&COL[nx][ny]==0)
							{
								Q.push(make_pair(nx,ny));
								if(s[nx][ny]=='#')cntB++;
								else cntW++;
								COL[nx][ny] = 1;
							}
						}
					}
				}
				ans += cntW*cntB;
			}

		}
	}
	cout<<ans<<endl;
	return 0;
}