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 dx[9] = {0, 1, -1, 0, 0, -1, -1, 1, 1};
const int dy[9] = {0, 0, 0, -1, 1, -1, 1, -1, 1};
const double pi = acos(-1.0);
const int N = 1e3;
struct abc {
  int d[6];
};
int n;
long long b[N + 10][6];
abc temp1, temp2;
vector<long long> v;
long long cal(abc temp1, abc temp2) {
  long long ju = 0;
  for (int i = 1; i <= 5; i++) ju += temp1.d[i] * temp2.d[i];
  return ju;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= 5; j++) scanf("%lld", &b[i][j]);
  if (n > 250) {
    puts("0");
  } else {
    for (int i = 1; i <= n; i++) {
      bool ok = true;
      for (int j = 1; j <= n; j++) {
        if (i == j) continue;
        if (!ok) break;
        for (int l = 1; l <= 5; l++) temp1.d[l] = b[j][l] - b[i][l];
        for (int k = 1; k <= n; k++) {
          if (j == k || i == k) continue;
          for (int l = 1; l <= 5; l++) {
            temp2.d[l] = b[k][l] - b[i][l];
          }
          if (cal(temp1, temp2) > 0) {
            ok = false;
            break;
          }
        }
      }
      if (ok) {
        v.emplace_back(i);
      }
    }
    printf("%d", ((int)v.size()));
    puts("");
    for (int i = 0; i <= ((int)v.size()) - 1; i++) {
      printf("%d", v[i]), puts("");
    }
  }
  return 0;
}
