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>
#define int long long
using namespace std;

char s[500][500];
bool vis[500][500];
int ans,n,m;

const int dx[4]={0,0,1,-1};
const int dy[4]={1,-1,0,0};

signed main()
{
	ios::sync_with_stdio(false);
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			cin>>s[i][j];
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++) if(!vis[i][j]){
			int cntb=0,cntw=0;
			vis[i][j]=true;
			if(s[i][j]=='#') cntb++;
			else cntw++;
			queue<pair<int,int> > q;
			q.push(make_pair(i,j));
			while(!q.empty()){
				pair<int,int> t=q.front();
				q.pop();
				for(int d=0;d<4;d++){
					int nx=t.first+dx[d],ny=t.second+dy[d];
					if(nx<=0||ny<=0||nx>n||ny>m) continue;
					if(vis[nx][ny]) continue;
					if(s[t.first][t.second]==s[nx][ny]) continue;
					vis[nx][ny]=true;
					q.push(make_pair(nx,ny));
					if(s[nx][ny]=='#') cntb++;
					else cntw++;
				}
			}
			ans+=cntb*cntw;
		}
	}
	cout<<ans<<endl;
	return 0;
}