#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
long long dot(const vector<int> &a, const vector<int> &b) {
  long long ans = 0;
  for (int i = 0; i < 5; i++) ans += a[i] * b[i];
  return ans;
}
long double abs(const vector<int> &a) { return sqrt(dot(a, a)); }
long double angle(const vector<int> &a, const vector<int> &b) {
  return dot(a, b) / (abs(a) * abs(b));
}
vector<int> operator-(const vector<int> &a, const vector<int> &b) {
  vector<int> ans(5);
  for (int i = 0; i < 5; i++) ans[i] = a[i] - b[i];
  return ans;
}
vector<int> project(int dim1, int dim2, const vector<int> &a) {
  vector<int> ans(5);
  ans[dim1] = a[dim1];
  ans[dim2] = a[dim2];
  return ans;
}
const long double PI = acos(-1.L);
const long double EPS = 1e-10;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> pts(n, vector<int>(5));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) cin >> pts[i][j];
  bitset<1001> good;
  for (int i = 0; i < 1001; i++) good[i] = 1;
  for (int i = 0; i < n; i++) {
    if (!good[i]) continue;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        long double b = angle(pts[k] - pts[i], pts[j] - pts[i]);
        long double c = angle(pts[i] - pts[j], pts[k] - pts[j]);
        if (b > 0) good[i] = 0;
        if (c > 0) good[j] = 0;
        if (PI - acos(b) - acos(c) < PI / 2 - EPS) good[k] = 0;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++)
    if (good[i]) ans++;
  cout << ans << '\n';
  for (int i = 0; i < n; i++)
    if (good[i]) cout << i + 1 << " ";
  cout << '\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;
const int MOD = 1e9 + 7;
long long dot(const vector<int> &a, const vector<int> &b) {
  long long ans = 0;
  for (int i = 0; i < 5; i++) ans += a[i] * b[i];
  return ans;
}
long double abs(const vector<int> &a) { return sqrt(dot(a, a)); }
long double angle(const vector<int> &a, const vector<int> &b) {
  return dot(a, b) / (abs(a) * abs(b));
}
vector<int> operator-(const vector<int> &a, const vector<int> &b) {
  vector<int> ans(5);
  for (int i = 0; i < 5; i++) ans[i] = a[i] - b[i];
  return ans;
}
vector<int> project(int dim1, int dim2, const vector<int> &a) {
  vector<int> ans(5);
  ans[dim1] = a[dim1];
  ans[dim2] = a[dim2];
  return ans;
}
const long double PI = acos(-1.L);
const long double EPS = 1e-10;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> pts(n, vector<int>(5));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) cin >> pts[i][j];
  bitset<1001> good;
  for (int i = 0; i < 1001; i++) good[i] = 1;
  for (int i = 0; i < n; i++) {
    if (!good[i]) continue;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        long double b = angle(pts[k] - pts[i], pts[j] - pts[i]);
        long double c = angle(pts[i] - pts[j], pts[k] - pts[j]);
        if (b > 0) good[i] = 0;
        if (c > 0) good[j] = 0;
        if (PI - acos(b) - acos(c) < PI / 2 - EPS) good[k] = 0;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++)
    if (good[i]) ans++;
  cout << ans << '\n';
  for (int i = 0; i < n; i++)
    if (good[i]) cout << i + 1 << " ";
  cout << '\n';
  return 0;
}

```