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;
char s[405][405];
bool S[405][405];
ll ans;
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
typedef pair<int,int> P;
int main(void){
    int a,b; cin>>a>>b;
    for(int i=0;i<a;i++){
		for(int j=0;j<b;j++){
			cin>>s[i][j];
		}
	}
	queue<P> que;
	for(int i=0;i<a;i++){
		for(int j=0;j<b;j++){
			if(s[i][j]=='#'&&S[i][j]==false){
				ll count=1;
				ll countans=0;
				que.push(P(i,j));
				S[i][j]=true;
				while(que.size()){
					P p=que.front(); que.pop();
					for(int k=0;k<4;k++){
						int nx=p.first+dx[k],ny=p.second+dy[k];
						if(nx>=0&&nx<a&&ny>=0&&ny<b&&S[nx][ny]==false){
							if(s[p.first][p.second]=='#'){
								if(s[nx][ny]=='.'){
									countans++;
									que.push(P(nx,ny));
									S[nx][ny]=true;
								}
							}else{
								if(s[nx][ny]=='#'){
									count++;
									que.push(P(nx,ny));
									S[nx][ny]=true;
								}
							}
						}
					}
				}
				ans=ans+count*countans;
			}
		}
	}
	cout<<ans<<endl;
    
}