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;
int n;
int a[1010], b[1010], c[1010], d[1010], e[1010];
bool bad[1010];
inline int f(const int &i, const int &j, const int &k) {
  return (a[j] - a[i]) * (a[k] - a[i]) + (b[j] - b[i]) * (b[k] - b[i]) +
         (c[j] - c[i]) * (c[k] - c[i]) + (d[j] - d[i]) * (d[k] - d[i]) +
         (e[j] - e[i]) * (e[k] - e[i]);
}
int main() {
  int i, j, k, cnt = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d%d%d%d", a + i, b + i, c + i, d + i, e + i);
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; !bad[i] && j < n; j++) {
      if (i == j) continue;
      for (k = j + 1; k <= n; k++) {
        if (k == i) continue;
        if (f(i, j, k) > 0) {
          bad[i] = true;
          break;
        }
      }
    }
    if (!bad[i]) ++cnt;
  }
  printf("%d\n", cnt);
  for (i = 1; i <= n; i++) {
    if (!bad[i]) {
      printf("%d\n", i);
    }
  }
}
