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;

int h,w;
int vis[405][405],g[405][405];
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
pair<int,int> bfs(int sx,int sy){
	int black=0,white=0;
	queue<pair<int,int> >q;
	q.push({sx,sy});
	vis[sx][sy]=1;
	while(!q.empty()){
		pair<int,int>f=q.front();q.pop();
		int x=f.first;
		int y=f.second;
		int p=g[x][y];
		if(p==0)black++;
		else white++;
		for(int i=0;i<4;i++){
			int xx=x+dx[i],yy=y+dy[i];
			if(xx>=0 and xx<h and yy>=0 and yy<w){
				int pp=g[xx][yy];
				if(1-pp==p and !vis[xx][yy]){
					vis[xx][yy]=1;
					q.push({xx,yy});
				}
			}
		}
	}
	return make_pair(black,white);
}

int main(){ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin>>h>>w;

for(int i=0;i<h;i++){
	for(int j=0;j<w;j++){
		char c;cin>>c;
		g[i][j]=(c=='.');
	}
}
long long ans=0;
for(int i=0;i<h;i++){
	for(int j=0;j<w;j++){
		if(!vis[i][j]){
			pair<int,int> f=bfs(i,j);
			ans+=1LL*f.first*f.second;
		}
	}
}
cout<<ans;
return 0;
}