#include <bits/stdc++.h>
using namespace std;
const int MAX = 1111;
struct Point {
  int a, b, c, d, e;
  Point() {}
};
int n;
Point p[MAX];
int dst[MAX][MAX];
bool bad[MAX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> p[i].a >> p[i].b >> p[i].c >> p[i].d >> p[i].e;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      dst[i][j] = p[i].a * p[j].a + p[i].b * p[j].b + p[i].c * p[j].c +
                  p[i].d * p[j].d + p[i].e * p[j].e;
    }
  }
  int ans = n;
  for (int k = 0; k < n; ++k) {
    for (int i = 0; i < n; ++i) {
      if (i == k) continue;
      for (int j = i + 1; j < n; ++j) {
        if (dst[i][j] - dst[k][i] - dst[k][j] + dst[k][k] > 0) bad[k] = true;
      }
    }
    if (bad[k]) --ans;
  }
  cout << ans << "\n";
  for (int i = 0; i < n; ++i) {
    if (!bad[i]) cout << i + 1 << "\n";
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 MAX = 1111;
struct Point {
  int a, b, c, d, e;
  Point() {}
};
int n;
Point p[MAX];
int dst[MAX][MAX];
bool bad[MAX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> p[i].a >> p[i].b >> p[i].c >> p[i].d >> p[i].e;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      dst[i][j] = p[i].a * p[j].a + p[i].b * p[j].b + p[i].c * p[j].c +
                  p[i].d * p[j].d + p[i].e * p[j].e;
    }
  }
  int ans = n;
  for (int k = 0; k < n; ++k) {
    for (int i = 0; i < n; ++i) {
      if (i == k) continue;
      for (int j = i + 1; j < n; ++j) {
        if (dst[i][j] - dst[k][i] - dst[k][j] + dst[k][k] > 0) bad[k] = true;
      }
    }
    if (bad[k]) --ans;
  }
  cout << ans << "\n";
  for (int i = 0; i < n; ++i) {
    if (!bad[i]) cout << i + 1 << "\n";
  }
  return 0;
}

```