#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int N = 1e2;
const double EPS = 1e-9;
const int INF = 1e9;
const double PI = acos(-1);
int n;
int ar[N + 5][5];
int sqr(int a) { return a * a; }
int dot(int idxi, int idxa, int idxb) {
  int ans = 0;
  for (int i = 0; i < 5; ++i) {
    ans += (ar[idxa][i] - ar[idxi][i]) * (ar[idxb][i] - ar[idxi][i]);
  }
  return ans;
}
double len(int idxi, int idxa) { return sqrt(dot(idxi, idxa, idxa)); }
double angle(int idxi, int idxa, int idxb) {
  return acos(1.0 * dot(idxi, idxa, idxb) / len(idxi, idxa) / len(idxi, idxb));
}
int main() {
  scanf("%d", &n);
  if (n >= N) {
    puts("0");
    return 0;
  }
  if (n <= 2) {
    printf("%d\n", n);
    for (int i = 1; i <= n; ++i) {
      printf("%d\n", i);
    }
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 5; ++j) {
      scanf("%d", &ar[i][j]);
    }
  }
  vector<int> v;
  for (int i = 0; i < n; ++i) {
    bool bisa = true;
    for (int j = 0; j < n && bisa; ++j) {
      if (i == j) continue;
      for (int k = j + 1; k < n && bisa; ++k) {
        if (k == i) continue;
        double angl = fabs(angle(i, j, k));
        if (PI - 2 * angl > EPS) {
          bisa = false;
        }
      }
    }
    if (bisa) v.push_back(i);
  }
  printf("%d\n", (int)v.size());
  for (int i = 0; i < v.size(); ++i) {
    printf("%d\n", v[i] + 1);
  }
  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;
const int MOD = 1e9 + 7;
const int N = 1e2;
const double EPS = 1e-9;
const int INF = 1e9;
const double PI = acos(-1);
int n;
int ar[N + 5][5];
int sqr(int a) { return a * a; }
int dot(int idxi, int idxa, int idxb) {
  int ans = 0;
  for (int i = 0; i < 5; ++i) {
    ans += (ar[idxa][i] - ar[idxi][i]) * (ar[idxb][i] - ar[idxi][i]);
  }
  return ans;
}
double len(int idxi, int idxa) { return sqrt(dot(idxi, idxa, idxa)); }
double angle(int idxi, int idxa, int idxb) {
  return acos(1.0 * dot(idxi, idxa, idxb) / len(idxi, idxa) / len(idxi, idxb));
}
int main() {
  scanf("%d", &n);
  if (n >= N) {
    puts("0");
    return 0;
  }
  if (n <= 2) {
    printf("%d\n", n);
    for (int i = 1; i <= n; ++i) {
      printf("%d\n", i);
    }
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 5; ++j) {
      scanf("%d", &ar[i][j]);
    }
  }
  vector<int> v;
  for (int i = 0; i < n; ++i) {
    bool bisa = true;
    for (int j = 0; j < n && bisa; ++j) {
      if (i == j) continue;
      for (int k = j + 1; k < n && bisa; ++k) {
        if (k == i) continue;
        double angl = fabs(angle(i, j, k));
        if (PI - 2 * angl > EPS) {
          bisa = false;
        }
      }
    }
    if (bisa) v.push_back(i);
  }
  printf("%d\n", (int)v.size());
  for (int i = 0; i < v.size(); ++i) {
    printf("%d\n", v[i] + 1);
  }
  return 0;
}

```