#include <bits/stdc++.h>
using namespace std;
double len[1010];
int a[1010][5];
bool bad[1010];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> a[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    if (bad[i]) continue;
    bool flag = false;
    for (int j = 0; j < n - 1; j++) {
      if (i == j) continue;
      if (flag) break;
      for (int k = j + 1; k < n; k++) {
        if (i == k) continue;
        double t = 0;
        double l1 = 0, l2 = 0;
        for (int u = 0; u < 5; u++) {
          t += (a[j][u] - a[i][u]) * (a[k][u] - a[i][u]);
          l1 += (a[j][u] - a[i][u]) * (a[j][u] - a[i][u]);
          l2 += (a[k][u] - a[i][u]) * (a[k][u] - a[i][u]);
        }
        t /= sqrt(l1);
        t /= sqrt(l2);
        if (t > 0 && t <= 1) {
          bad[i] = true;
          break;
          flag = true;
        } else {
          bad[j] = true;
          bad[k] = true;
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (!bad[i]) {
      ans++;
    }
  }
  cout << ans << endl;
  for (int i = 0; i < n; i++) {
    if (!bad[i]) {
      cout << i + 1 << endl;
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
double len[1010];
int a[1010][5];
bool bad[1010];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      cin >> a[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    if (bad[i]) continue;
    bool flag = false;
    for (int j = 0; j < n - 1; j++) {
      if (i == j) continue;
      if (flag) break;
      for (int k = j + 1; k < n; k++) {
        if (i == k) continue;
        double t = 0;
        double l1 = 0, l2 = 0;
        for (int u = 0; u < 5; u++) {
          t += (a[j][u] - a[i][u]) * (a[k][u] - a[i][u]);
          l1 += (a[j][u] - a[i][u]) * (a[j][u] - a[i][u]);
          l2 += (a[k][u] - a[i][u]) * (a[k][u] - a[i][u]);
        }
        t /= sqrt(l1);
        t /= sqrt(l2);
        if (t > 0 && t <= 1) {
          bad[i] = true;
          break;
          flag = true;
        } else {
          bad[j] = true;
          bad[k] = true;
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (!bad[i]) {
      ans++;
    }
  }
  cout << ans << endl;
  for (int i = 0; i < n; i++) {
    if (!bad[i]) {
      cout << i + 1 << endl;
    }
  }
  return 0;
}

```