#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool scan_d(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
struct mmp {
  int a, b, c, d, e;
} p[1005];
int ans[1005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++)
    cin >> p[i].a >> p[i].b >> p[i].c >> p[i].d >> p[i].e;
  int cnt = 0;
  for (int k = 1; k <= n; k++) {
    bool flag = true;
    for (int i = 1; i <= n && flag; i++) {
      if (i == k) continue;
      for (int j = i + 1; j <= n; j++) {
        if (j == k) continue;
        int tmp = (p[j].a - p[k].a) * (p[i].a - p[k].a) +
                  (p[j].b - p[k].b) * (p[i].b - p[k].b) +
                  (p[j].c - p[k].c) * (p[i].c - p[k].c) +
                  (p[j].d - p[k].d) * (p[i].d - p[k].d) +
                  (p[j].e - p[k].e) * (p[i].e - p[k].e);
        if (tmp > 0) {
          flag = false;
          break;
        }
      }
    }
    if (flag) ans[++cnt] = k;
  }
  cout << cnt << "\n";
  for (int i = 1; i <= cnt; i++) cout << ans[i] << "\n";
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
template <class T>
inline bool scan_d(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
struct mmp {
  int a, b, c, d, e;
} p[1005];
int ans[1005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++)
    cin >> p[i].a >> p[i].b >> p[i].c >> p[i].d >> p[i].e;
  int cnt = 0;
  for (int k = 1; k <= n; k++) {
    bool flag = true;
    for (int i = 1; i <= n && flag; i++) {
      if (i == k) continue;
      for (int j = i + 1; j <= n; j++) {
        if (j == k) continue;
        int tmp = (p[j].a - p[k].a) * (p[i].a - p[k].a) +
                  (p[j].b - p[k].b) * (p[i].b - p[k].b) +
                  (p[j].c - p[k].c) * (p[i].c - p[k].c) +
                  (p[j].d - p[k].d) * (p[i].d - p[k].d) +
                  (p[j].e - p[k].e) * (p[i].e - p[k].e);
        if (tmp > 0) {
          flag = false;
          break;
        }
      }
    }
    if (flag) ans[++cnt] = k;
  }
  cout << cnt << "\n";
  for (int i = 1; i <= cnt; i++) cout << ans[i] << "\n";
  return 0;
}

```