#include <bits/stdc++.h>
using namespace std;
int n;
struct Point {
  int a, b, c, d, e;
  Point() {}
  Point(int _a, int _b, int _c, int _d, int _e)
      : a(_a), b(_b), c(_c), d(_d), e(_e) {}
  void Read() { scanf("%d%d%d%d%d", &a, &b, &c, &d, &e); }
  friend Point operator-(const Point &x, const Point &y) {
    return Point(x.a - y.a, x.b - y.b, x.c - y.c, x.d - y.d, x.e - y.e);
  }
  friend int operator*(const Point &x, const Point &y) {
    return x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e;
  }
} p[1005];
bool Judge(int i, int j, int k) {
  long long t1 = (p[j] - p[i]) * (p[k] - p[i]);
  if (t1 <= 0) return 0;
  t1 *= t1;
  long long t2 = (p[j] - p[i]) * (p[j] - p[i]);
  t2 *= (p[k] - p[i]) * (p[k] - p[i]);
  return 0 < t1 && t1 <= t2;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    p[i].Read();
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    bool flag = 1;
    for (int j = 1; flag && j <= n; j++) {
      if (j == i) continue;
      for (int k = j + 1; flag && k <= n; k++) {
        if (k == i) continue;
        if (Judge(i, j, k)) {
          flag = 0;
          break;
        }
      }
    }
    if (flag) {
      ans.push_back(i);
    }
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < int(ans.size()); i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### Prompt

Generate 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;
int n;
struct Point {
  int a, b, c, d, e;
  Point() {}
  Point(int _a, int _b, int _c, int _d, int _e)
      : a(_a), b(_b), c(_c), d(_d), e(_e) {}
  void Read() { scanf("%d%d%d%d%d", &a, &b, &c, &d, &e); }
  friend Point operator-(const Point &x, const Point &y) {
    return Point(x.a - y.a, x.b - y.b, x.c - y.c, x.d - y.d, x.e - y.e);
  }
  friend int operator*(const Point &x, const Point &y) {
    return x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e;
  }
} p[1005];
bool Judge(int i, int j, int k) {
  long long t1 = (p[j] - p[i]) * (p[k] - p[i]);
  if (t1 <= 0) return 0;
  t1 *= t1;
  long long t2 = (p[j] - p[i]) * (p[j] - p[i]);
  t2 *= (p[k] - p[i]) * (p[k] - p[i]);
  return 0 < t1 && t1 <= t2;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    p[i].Read();
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    bool flag = 1;
    for (int j = 1; flag && j <= n; j++) {
      if (j == i) continue;
      for (int k = j + 1; flag && k <= n; k++) {
        if (k == i) continue;
        if (Judge(i, j, k)) {
          flag = 0;
          break;
        }
      }
    }
    if (flag) {
      ans.push_back(i);
    }
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < int(ans.size()); i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```