#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e4 + 5;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
template <class T>
inline bool read(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;
}
vector<int> v[1005];
int f(int p, int q, int r) {
  int sum = 0;
  for (int i = 0; i < 5; ++i) {
    sum += (v[p][i] - v[q][i]) * (v[p][i] - v[r][i]);
  }
  return sum;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x;
    for (int j = 0; j < 5; ++j) {
      cin >> x;
      v[i].push_back(x);
    }
  }
  if (n > 11)
    printf("0");
  else {
    vector<int> ans;
    for (int i = 1; i <= n; ++i) {
      int flag = 1;
      for (int j = 1; j < n; ++j) {
        for (int k = j + 1; k <= n; ++k) {
          if (i == j || i == k) continue;
          if (f(i, j, k) > 0) {
            flag = 0;
            break;
          }
        }
        if (flag == 0) break;
      }
      if (flag) ans.push_back(i);
    }
    printf("%d\n", ans.size());
    for (int i = 0; i < ans.size(); ++i) printf("%d ", ans[i]);
  }
  return 0;
}

### Prompt

Your task is to create 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 int MAXN = 5e4 + 5;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
template <class T>
inline bool read(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;
}
vector<int> v[1005];
int f(int p, int q, int r) {
  int sum = 0;
  for (int i = 0; i < 5; ++i) {
    sum += (v[p][i] - v[q][i]) * (v[p][i] - v[r][i]);
  }
  return sum;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x;
    for (int j = 0; j < 5; ++j) {
      cin >> x;
      v[i].push_back(x);
    }
  }
  if (n > 11)
    printf("0");
  else {
    vector<int> ans;
    for (int i = 1; i <= n; ++i) {
      int flag = 1;
      for (int j = 1; j < n; ++j) {
        for (int k = j + 1; k <= n; ++k) {
          if (i == j || i == k) continue;
          if (f(i, j, k) > 0) {
            flag = 0;
            break;
          }
        }
        if (flag == 0) break;
      }
      if (flag) ans.push_back(i);
    }
    printf("%d\n", ans.size());
    for (int i = 0; i < ans.size(); ++i) printf("%d ", ans[i]);
  }
  return 0;
}

```