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.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

public class C {

	public static void main(String[] args) {
		try (Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
				PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)))) {
			final int n = in.nextInt();
			final int[][] a = new int[n][5];
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < 5; j++) {
					a[i][j] = in.nextInt();
				}
			}
			List<Integer> ans = solve(a);
			out.println(ans.size());
			for (Integer i : ans) {
				out.println(i);
			}
		}
	}

	private static List<Integer> solve(int[][] a) {
		final List<Integer> ans = new LinkedList<>();
		final int n = a.length;
		if (n < 3) {
			ans.add(1);
			if (n > 1) {
				ans.add(2);
			}
			return ans;
		}
		final double d = Math.PI / 2 - 1e-9;
		points: for (int i = 0; i < n; i++) {
			for (int k = 0; k < n; k++) {
				if (k == i) {
					continue;
				}
				for (int j = 0; j < n; j++) {
					if (j != i && j != k) {
						if (Math.acos(((a[k][0] - a[i][0]) * (a[j][0] - a[i][0])
								+ (a[k][1] - a[i][1]) * (a[j][1] - a[i][1]) + (a[k][2] - a[i][2]) * (a[j][2] - a[i][2])
								+ (a[k][3] - a[i][3]) * (a[j][3] - a[i][3]) + (a[k][4] - a[i][4]) * (a[j][4] - a[i][4]))
								/ Math.sqrt((a[k][0] - a[i][0]) * (a[k][0] - a[i][0])
										+ (a[k][1] - a[i][1]) * (a[k][1] - a[i][1])
										+ (a[k][2] - a[i][2]) * (a[k][2] - a[i][2])
										+ (a[k][3] - a[i][3]) * (a[k][3] - a[i][3])
										+ (a[k][4] - a[i][4]) * (a[k][4] - a[i][4]))
								/ Math.sqrt((a[j][0] - a[i][0]) * (a[j][0] - a[i][0])
										+ (a[j][1] - a[i][1]) * (a[j][1] - a[i][1])
										+ (a[j][2] - a[i][2]) * (a[j][2] - a[i][2])
										+ (a[j][3] - a[i][3]) * (a[j][3] - a[i][3])
										+ (a[j][4] - a[i][4]) * (a[j][4] - a[i][4]))) < d) {
							continue points;
						}
					}
				}
			}
			ans.add(i + 1);
		}
		return ans;
	}

}
