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;


public class CF_8C {

	public static void main(String[] args) {
		
		// Hooray bitmasks! I'm good at these :)
		
		Scanner in = new Scanner(System.in);
		
		// Handbag coordinates..
		int hb_x = in.nextInt(), hb_y = in.nextInt();
		int n = in.nextInt();
		int[] ox = new int[n];
		int[] oy = new int[n];
		
		// Dynamic programming: Also store a matrix of the time to reach one
		//  object from the other.
		// This considers the handbag to be object 0
		int[][] dt = new int[n][n];
		int[] hbd = new int[n];
		for (int i = 0; i < n; i++) {
			ox[i] = in.nextInt();
			oy[i] = in.nextInt();
			hbd[i] = (ox[i] - hb_x) * (ox[i] - hb_x)
					+ (oy[i] - hb_y) * (oy[i] - hb_y);
		}
		
		// Compute elapsed times...
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				dt[i][j] = (ox[i] - ox[j]) * (ox[i] - ox[j])
						+ (oy[i] - oy[j]) * (oy[i] - oy[j]);
			}
		}
		
		// Fill up an array with the amount of time it takes to grab
		//  all objects with the specified bitmask.
		int[] sofar = new int[1 << n];
		int[] masks = new int[1 << n];
		sofar[0] = 0;
		for (int i = 1; i < (1 << n); i++) {
			sofar[i] = -1;
		}
		
		for (int i = 0; i < (1 << n); i++) {
			if (sofar[i] != -1) {
				for (int maskbit = 0; maskbit < n; maskbit++) {
					// Look for first object in bitmask to grab...
					if (((1 << maskbit) & i) == 0) {
						int iffirst = ((1 << maskbit) | i);
						int fromold = sofar[i] + 2 * hbd[maskbit];
						
						if (sofar[iffirst] == -1 || sofar[iffirst] > fromold) {
							// A better way to get to position J was found, use it.
							sofar[iffirst] = fromold;
							masks[iffirst] = i;
						}
						
						// Find another thing while you're out...
						for (int otherone = 0; otherone < n; otherone++) {
							if (((1 << otherone) & iffirst) == 0) {
								int iffollow = ((1 << otherone) | iffirst);
								int fromi = sofar[i] + hbd[maskbit] + dt[maskbit][otherone] + hbd[otherone];
								
								// Did we find a better way to get to iffollow state?
								if (sofar[iffollow] == -1 || sofar[iffollow] > fromi) {
									sofar[iffollow] = fromi;
									masks[iffollow] = i;
								}
							}
						}
						break;
					}
				}
			}
		}
		
		// After all this time, we have an answer.
		// The minimum time will be the value of sofar at the very end,
		//  which will have the case of if all objects were picked up.
		// Clever, no?
		// The logic came from http://www.darrensun.com/codeforces-round-8/
		// Wish I could claim it mine, but it is not so.
		
		int end_val = (1 << n) - 1;
		
		System.out.println(sofar[end_val]);
		System.out.print(0);
		while (end_val > 0) {
			// Which objects were collected in the prvious trip?
			int diff = end_val ^ masks[end_val];
			int obj1 = -1, obj2 = -1;
			for (int i = 0; i < n; i++) {
				if (((1 << i) & diff) > 0) {
					obj2 = obj1;
					obj1 = i;
				}
			}
			
			if (obj2 >= 0) {
				// Two objects were collected this trip, output them both.
				System.out.print(" " + (obj1 + 1) + " " + (obj2 + 1) + " 0");
			} else {
				// Only one object was collected here.
				System.out.print(" " + (obj1 + 1) + " 0");
			}
			end_val = masks[end_val];
		}
		
		in.close();
	}

}