You are given set of n points in 5-dimensional space. The points are labeled from 1 to n. No two points coincide.

We will call point a bad if there are different points b and c, not equal to a, from the given set such that angle between vectors <image> and <image> is acute (i.e. strictly less than <image>). Otherwise, the point is called good.

The angle between vectors <image> and <image> in 5-dimensional space is defined as <image>, where <image> is the scalar product and <image> is length of <image>.

Given the list of points, print the indices of the good points in ascending order.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 103) — the number of points.

The next n lines of input contain five integers ai, bi, ci, di, ei (|ai|, |bi|, |ci|, |di|, |ei| ≤ 103) — the coordinates of the i-th point. All points are distinct.

Output

First, print a single integer k — the number of good points.

Then, print k integers, each on their own line — the indices of the good points in ascending order.

Examples

Input

6
0 0 0 0 0
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1


Output

1
1


Input

3
0 0 1 2 0
0 0 9 2 0
0 0 5 9 0


Output

0

Note

In the first sample, the first point forms exactly a <image> angle with all other pairs of points, so it is good.

In the second sample, along the cd plane, we can see the points look as follows:

<image>

We can see that all angles here are acute, so no points are good.
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Scanner;

/**
 * Created by Pranav ps on 04-09-2017.
 */
public class wedss {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        if(n <= 11){
            long[][] points = new long[n][5];

            for(int i = 0 ; i < n; i++){
                for (int j = 0; j < 5; j++) {
                    points[i][j] = sc.nextLong();
                }
            }

            long[][][] vectors = new long[n][n][5];
            for(int i = 0; i < n; i++){
                for(int j = 0; j < n; j++){
                    for (int k = 0; k < 5; k++) {
                        vectors[i][j][k] = points[j][k] - points[i][k];
                    }
                }
            }

            HashSet<Integer> list = new HashSet<>();

            for(int i = 0; i < n; i++){
                boolean flag = false;
                for (int j = 0; j < n; j++) {
                    for (int k = 0; k < n; k++) {
                        if(!equal(vectors[i][j],vectors[i][k]) && dot(vectors[i][j],vectors[i][k])){
                            list.add(i+1);
                            flag = true;
                            break;
                        }
                    }
                    if(flag)
                        break;
                }
            }

            System.out.println(n - list.size());
            for(int i = 0; i < n; i++){
                if(!list.contains(i+1))
                    System.out.println(i+1);
            }
        }
        else{
            System.out.println(0);
        }
    }
    private static boolean equal(long[] a, long[] b){
        boolean res = true;
        for(int i = 0 ; i < 5; i++){
            res &= a[i]==b[i];
        }
        return res;
    }
    private static boolean dot(long[] a, long[] b){
        long res = 0;
        for(int i = 0 ; i < 5; i++){
            res += a[i]*b[i];
        }

        return res>0;
    }
}