#include <bits/stdc++.h>
using namespace std;
const int N = 1007;
const int MX = 1e9 + 7;
const int inf = 1e9 + 9;
int n, m;
int tab[N][5];
int main() {
  scanf("%d", &n);
  if (n > 33) {
    puts("0");
    return 0;
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < 5; ++j) scanf("%d", &tab[i][j]);
  vector<int> res;
  for (int i = 1; i <= n; ++i) {
    bool ans = true;
    for (int j = 1; j <= n; ++j)
      for (int k = 1; k <= n; ++k) {
        if (i == j || j == k || i == k) continue;
        int cur = 0;
        for (int ii = 0; ii < 5; ++ii)
          cur += (tab[j][ii] - tab[i][ii]) * (tab[k][ii] - tab[i][ii]);
        if (cur > 0) ans = false;
      }
    if (ans) res.push_back(i);
  }
  printf("%d\n", res.size());
  for (int v : res) printf("%d ", v);
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1007;
const int MX = 1e9 + 7;
const int inf = 1e9 + 9;
int n, m;
int tab[N][5];
int main() {
  scanf("%d", &n);
  if (n > 33) {
    puts("0");
    return 0;
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < 5; ++j) scanf("%d", &tab[i][j]);
  vector<int> res;
  for (int i = 1; i <= n; ++i) {
    bool ans = true;
    for (int j = 1; j <= n; ++j)
      for (int k = 1; k <= n; ++k) {
        if (i == j || j == k || i == k) continue;
        int cur = 0;
        for (int ii = 0; ii < 5; ++ii)
          cur += (tab[j][ii] - tab[i][ii]) * (tab[k][ii] - tab[i][ii]);
        if (cur > 0) ans = false;
      }
    if (ans) res.push_back(i);
  }
  printf("%d\n", res.size());
  for (int v : res) printf("%d ", v);
  return 0;
}

```