Problem: Girl Lena likes it when everything is in order, and looks for order everywhere. Once she was getting ready for the University and noticed that the room was in a mess — all the objects from her handbag were thrown about the room. Of course, she wanted to put them back into her handbag. The problem is that the girl cannot carry more than two objects at a time, and cannot move the handbag. Also, if he has taken an object, she cannot put it anywhere except her handbag — her inherent sense of order does not let her do so.

You are given the coordinates of the handbag and the coordinates of the objects in some Сartesian coordinate system. It is known that the girl covers the distance between any two objects in the time equal to the squared length of the segment between the points of the objects. It is also known that initially the coordinates of the girl and the handbag are the same. You are asked to find such an order of actions, that the girl can put all the objects back into her handbag in a minimum time period.

Input

The first line of the input file contains the handbag's coordinates xs, ys. The second line contains number n (1 ≤ n ≤ 24) — the amount of objects the girl has. The following n lines contain the objects' coordinates. All the coordinates do not exceed 100 in absolute value. All the given positions are different. All the numbers are integer.

Output

In the first line output the only number — the minimum time the girl needs to put the objects into her handbag. 

In the second line output the possible optimum way for Lena. Each object in the input is described by its index number (from 1 to n), the handbag's point is described by number 0. The path should start and end in the handbag's point. If there are several optimal paths, print any of them. 

Examples

Input

0 0
2
1 1
-1 1


Output

8
0 1 2 0 


Input

1 1
3
4 3
3 4
0 0


Output

32
0 1 2 0 3 0 

Solution (in JAVA):

Solution:
import java.util.Scanner;
import java.io.PrintWriter;
import java.util.*;


public class Order8C implements Runnable {
    private Scanner in = new Scanner(System.in);
    private PrintWriter out = new PrintWriter(System.out);

    int mintime;
    String path;

    int xs;
    int ys;
    int n;
    int[] obx;
    int[] oby;


    public static void main(String[] args) {
	new Thread(new Order8C()).start();
    }

    private void read() {
	xs = in.nextInt();
	ys = in.nextInt();
	n = in.nextInt();
	obx = new int[n];
	oby = new int[n];
	for(int i = 0; i<n; i++){
	    obx[i] = in.nextInt();
	    oby[i] = in.nextInt();
	}
	
    }

    private void solve() {
	/* we will do dynamic programming over the subsets
	   best[m] will be the shortest for binary digits of m
	   being in. 
	   last[m] will be the last subset before m */

	int[] ds = new int[n];
	int[][] d = new int[n][n];

	int[] best = new int[1 << n];
	int[] last = new int[1 << n];


	for(int i = 0; i<n; i++){
	    ds[i] = (obx[i]-xs)*(obx[i]-xs) + (oby[i]-ys)*(oby[i]-ys);
	    for(int j = i+1; j<n; j++){
		d[i][j] = (obx[i]-obx[j])*(obx[i]-obx[j]) + (oby[i]-oby[j])*(oby[i]-oby[j]);
		d[j][i] = (obx[i]-obx[j])*(obx[i]-obx[j]) + (oby[i]-oby[j])*(oby[i]-oby[j]);
	    }
	}


	for(int i=0; i< (1<<n); i++){ //goes through subsets before supersets
	    best[i] = 100000000;
	}
	best[0] = 0;

	for(int i=0; i< (1<<n); i++){ //goes through subsets before supersets
	    /*if (i % 1000000 == 0 ){
		System.out.println(i);
		}*/
	    for(int j = 0; j<n; j++){
		if( ((1 << j) & i) != 0){ // if 2^j and i have bitwise sum nonzero
		    if(best[i - (1<<j)] + 2*ds[j] < best[i]){
			best[i] = best[i - (1<<j)] + 2*ds[j];
			last[i] = i - (1<< j);

		    }
		   for(int k = j+1; k<n; k++){
		       if( ((1 << k) & i) != 0){ // if 2^k and i have bitwise sum nonzero
			   if( (best[i-(1<<j) -(1<<k)] +ds[j]+ds[k]+d[j][k]) < best[i]){
			       best[i] = (best[i-(1<<j) -(1<<k)] +ds[j]+ds[k]+d[j][k]);
			       last[i] = i -(1<<j) - (1<< k);

			   }

		       }
		   }
		   break; // break here because order doesn't really matter, just knowing which pairs to roup
		}
	    }
	}
	int i = (1<<n) -1;
	mintime = best[i];
	path = "";
	
	while(i >0){
	    //System.out.println("Got to i = " + i + " and path is: " + path);
	    path = path + "0 ";
	    int dif = i-last[i]; // this has one or two 1s in binary
	    for(int j=0; j<n; j++){
		if( ((1<<j) & dif) != 0){ // if dif has a 1 in the j'th digit
		    path  = path + (j+1) + " ";
		}
	    }
	    
	    i = last[i];
	}
	path = path + "0";
    }

    private void write() {
	out.println(mintime);
	out.println(path);
    }
    public void run() {
	read();
	solve();
	write();
	out.close();
    }
}