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 long double drept = acos(0);
vector<int> ans;
struct point {
  long double x1, x2, x3, x4, x5;
};
point A[1005];
bool solve(int i, int j, int k) {
  point ij;
  ij.x1 = (A[i].x1 - A[j].x1);
  ij.x2 = (A[i].x2 - A[j].x2);
  ij.x3 = (A[i].x3 - A[j].x3);
  ij.x4 = (A[i].x4 - A[j].x4);
  ij.x5 = (A[i].x5 - A[j].x5);
  point ik;
  ik.x1 = (A[i].x1 - A[k].x1);
  ik.x2 = (A[i].x2 - A[k].x2);
  ik.x3 = (A[i].x3 - A[k].x3);
  ik.x4 = (A[i].x4 - A[k].x4);
  ik.x5 = (A[i].x5 - A[k].x5);
  long double xy = ij.x1 * ik.x1 + ij.x2 * ik.x2 + ij.x3 * ik.x3 +
                   ij.x4 * ik.x4 + ij.x5 * ik.x5;
  long double l1 = sqrt(ij.x1 * ij.x1 + ij.x2 * ij.x2 + ij.x3 * ij.x3 +
                        ij.x4 * ij.x4 + ij.x5 * ij.x5);
  long double l2 = sqrt(ik.x1 * ik.x1 + ik.x2 * ik.x2 + ik.x3 * ik.x3 +
                        ik.x4 * ik.x4 + ik.x5 * ik.x5);
  long double x_y = l1 * l2;
  if (acos(xy / x_y) < drept) {
    return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i].x1;
    cin >> A[i].x2;
    cin >> A[i].x3;
    cin >> A[i].x4;
    cin >> A[i].x5;
  }
  int contans = 0;
  for (int i = 1; i <= n; i++) {
    bool ok = true;
    for (int j = 1; j <= n; j++) {
      if (j == i) {
        continue;
      }
      for (int k = 1; k <= n; k++) {
        if (k == j || k == i) {
          continue;
        }
        if (solve(i, j, k)) {
          ok = false;
          break;
        }
      }
      if (ok == false) {
        break;
      }
    }
    if (ok) {
      contans++;
      ans.push_back(i);
    }
  }
  cout << contans << '\n';
  for (auto x : ans) {
    cout << x << '\n';
  }
  return 0;
}
