#include <bits/stdc++.h>
using namespace std;
const long long mod = 10e6;
const double pi = acos(-1);
struct point5 {
  void readin() { scanf("%d %d %d %d %d", p, p + 1, p + 2, p + 3, p + 4); }
  void neg() {
    p[0] = ~p[0] + 1;
    p[1] = ~p[1] + 1;
    p[2] = ~p[2] + 1;
    p[3] = ~p[3] + 1;
    p[4] = ~p[4] + 1;
  }
  friend point5 operator-(const point5& a, const point5& b) {
    point5 res(a);
    res.p[0] -= b.p[0];
    res.p[1] -= b.p[1];
    res.p[2] -= b.p[2];
    res.p[3] -= b.p[3];
    res.p[4] -= b.p[4];
    return res;
  }
  int dot(const point5& a) {
    return p[0] * a.p[0] + p[1] * a.p[1] + p[2] * a.p[2] + p[3] * a.p[3] +
           p[4] * a.p[4];
  }
  int p[5];
};
int n, t;
point5 a[1002];
bool bisa[1002];
int main() {
  memset(bisa, true, sizeof bisa);
  scanf("%d", &n);
  for (int i = 0; i < n; i++) a[i].readin();
  for (int i = 0; i < n; i++) {
    if (!bisa[i]) continue;
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        point5 b = a[j] - a[i];
        point5 c = a[k] - a[i];
        if (b.dot(c) > 0) {
          bisa[i] = false;
          break;
        } else {
          bisa[j] = bisa[k] = false;
        }
      }
      if (!bisa[i]) break;
    }
  }
  vector<int> v;
  for (int i = 0; i < n; i++) {
    if (bisa[i]) v.push_back(i + 1);
  }
  printf("%d\n", v.size());
  for (int& a : v) printf("%d\n", a);
  return 0;
}

### Prompt

Please formulate 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 long long mod = 10e6;
const double pi = acos(-1);
struct point5 {
  void readin() { scanf("%d %d %d %d %d", p, p + 1, p + 2, p + 3, p + 4); }
  void neg() {
    p[0] = ~p[0] + 1;
    p[1] = ~p[1] + 1;
    p[2] = ~p[2] + 1;
    p[3] = ~p[3] + 1;
    p[4] = ~p[4] + 1;
  }
  friend point5 operator-(const point5& a, const point5& b) {
    point5 res(a);
    res.p[0] -= b.p[0];
    res.p[1] -= b.p[1];
    res.p[2] -= b.p[2];
    res.p[3] -= b.p[3];
    res.p[4] -= b.p[4];
    return res;
  }
  int dot(const point5& a) {
    return p[0] * a.p[0] + p[1] * a.p[1] + p[2] * a.p[2] + p[3] * a.p[3] +
           p[4] * a.p[4];
  }
  int p[5];
};
int n, t;
point5 a[1002];
bool bisa[1002];
int main() {
  memset(bisa, true, sizeof bisa);
  scanf("%d", &n);
  for (int i = 0; i < n; i++) a[i].readin();
  for (int i = 0; i < n; i++) {
    if (!bisa[i]) continue;
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        point5 b = a[j] - a[i];
        point5 c = a[k] - a[i];
        if (b.dot(c) > 0) {
          bisa[i] = false;
          break;
        } else {
          bisa[j] = bisa[k] = false;
        }
      }
      if (!bisa[i]) break;
    }
  }
  vector<int> v;
  for (int i = 0; i < n; i++) {
    if (bisa[i]) v.push_back(i + 1);
  }
  printf("%d\n", v.size());
  for (int& a : v) printf("%d\n", a);
  return 0;
}

```