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, A[1005][10], X[10], Y[10];
int main() {
  cin >> N;
  for (int i = (1); i <= (N); i++)
    for (int j = (1); j <= (5); j++) cin >> A[i][j];
  vector<int> ans;
  for (int i = (1); i <= (N); i++) {
    bool good = true;
    for (int j = (1); j <= (N); j++)
      for (int k = (1); k <= (N); k++)
        if (i != j && i != k && j != k) {
          for (int z = (1); z <= (5); z++) X[z] = A[j][z] - A[i][z];
          for (int z = (1); z <= (5); z++) Y[z] = A[k][z] - A[i][z];
          int val = 0;
          for (int z = (1); z <= (5); z++) val += X[z] * Y[z];
          if (val > 0) good = false;
          if (!good) j = N + 1, k = N + 1;
        }
    if (good) ans.push_back(i);
  }
  cout << ans.size() << "\n";
  for (int x : ans) cout << x << "\n";
}
