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 long int Maxn3 = 1e3 + 10;
const long long int Maxn4 = 1e4 + 10;
const long long int Maxn5 = 1e5 + 10;
const long long int Maxn6 = 1e6 + 10;
const long long int Maxn7 = 1e7 + 10;
const long long int Maxn8 = 1e8 + 10;
const long long int Maxn9 = 1e9 + 10;
const long long int Maxn18 = 1e18 + 10;
const long long int Mod1 = 1e7 + 7;
const long long int Mod2 = 1e9 + 7;
const long long int LLMax = LLONG_MAX;
const long long int LLMin = LLONG_MIN;
const long long int INTMax = INT_MAX;
const long long int INTMin = INT_MIN;
long long int mn = LLMax, mx = LLMin;
long long int a[1010][5], ans[5], ansSz;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n; i++)
    for (long long int j = 0; j < 5; j++) cin >> a[i][j];
  if (n > 100) return cout << 0, 0;
  for (long long int i = 0; i < n; i++) {
    bool ok = 1;
    for (long long int j = 0; j < n; j++)
      for (long long int h = j + 1; h < n; h++) {
        if (j == i || h == i) continue;
        long long int v = 0;
        for (long long int k = 0; k < 5; k++)
          v += (a[j][k] - a[i][k]) * (a[h][k] - a[i][k]);
        ok &= v <= 0;
      }
    if (ok) ans[ansSz++] = i + 1;
  }
  cout << ansSz << '\n';
  for (long long int i = 0; i < ansSz; i++) cout << ans[i] << ' ';
  return 0;
}
