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.
#include <bits/stdc++.h>
using namespace std;
const int mixx = 1e3 + 5;
int a[mixx][5], n;
double len(int i, int j) {
  double now = 0;
  int k;
  for (k = 0; k < 5; k++) now += (a[j][k] - a[i][k]) * (a[j][k] - a[i][k]);
  return sqrt(now);
}
bool check(int i, int j, int k) {
  double now = 0;
  int p;
  for (p = 0; p < 5; p++) now += (a[j][p] - a[i][p]) * (a[k][p] - a[i][p]);
  double L = len(i, j) * len(i, k);
  return acos(now / L) < acos(-1.0) / 2;
}
int main() {
  scanf("%d", &n);
  int i, j, k;
  for (i = 1; i <= n; i++) {
    for (j = 0; j < 5; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  if (n > 243) {
    printf("%d\n", 0);
    return 0;
  }
  queue<int> q;
  for (i = 1; i <= n; i++) {
    int flag = 1;
    for (j = 1; j <= n && flag; j++) {
      if (i == j) continue;
      for (k = j + 1; k <= n && flag; k++) {
        if (i == k) continue;
        if (check(i, j, k)) flag = 0;
      }
    }
    if (flag) q.push(i);
  }
  printf("%d\n", q.size());
  while (!q.empty()) {
    printf("%d\n", q.front());
    q.pop();
  }
  return 0;
}
