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;
vector<int> ans;
int n;
int a[1001][6];
int b[1001][1001][6];
bool test(int i) {
  for (int i1 = 1; i1 <= n; i1++)
    for (int i2 = 1; i2 <= n; i2++)
      if (i != i1 and i != i2 and i1 != i2) {
        int s = 0;
        for (int k = 1; k <= 5; k++) s += b[i][i1][k] * b[i][i2][k];
        if (s > 0) return 0;
      }
  return 1;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= 5; j++) cin >> a[i][j];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      if (i != j) {
        for (int k = 1; k <= 5; k++) b[i][j][k] = a[j][k] - a[i][k];
      }
    }
  for (int i = 1; i <= n; i++)
    if (test(i)) ans.push_back(i);
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << endl;
  return 0;
}
