#include <bits/stdc++.h>
using namespace std;
const long long inf = 2147483647;
long long n;
long long bb[1111][11];
long long a[1111][11];
long long dian[1111][1111];
long long mo[1111];
long long cheng[1111][1111];
vector<long long> ans;
int main() {
  cin >> n;
  if (n > 100) {
    cout << 0 << endl;
    return 0;
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= 5; j++) cin >> bb[i][j];
  }
  for (long long t = 1; t <= n; t++) {
    memset(a, 0, sizeof(a));
    memset(dian, 0, sizeof(dian));
    memset(mo, 0, sizeof(mo));
    memset(cheng, 0, sizeof(cheng));
    long long num = 0;
    for (long long i = 1; i <= n; i++) {
      if (i == t) continue;
      num++;
      for (long long j = 1; j <= 5; j++) {
        a[num][j] = bb[i][j] - bb[t][j];
      }
    }
    for (long long i = 1; i <= num; i++) {
      for (long long j = i + 1; j <= num; j++) {
        for (long long k = 1; k <= 5; k++) {
          dian[i][j] += a[i][k] * a[j][k];
        }
        dian[j][i] = dian[i][j];
      }
    }
    for (long long i = 1; i <= num; i++) {
      for (long long j = 1; j <= 5; j++) {
        mo[i] += a[i][j] * a[i][j];
      }
    }
    for (long long i = 1; i <= num; i++) {
      for (long long j = i + 1; j <= num; j++) {
        cheng[i][j] = mo[i] * mo[j];
        cheng[j][i] = cheng[i][j];
      }
    }
    bool flag = 1;
    for (long long i = 1; i <= num; i++) {
      for (long long j = 1; j <= num; j++) {
        if (i == j) continue;
        if (dian[i][j] > 0 && dian[i][j] * dian[i][j] <= cheng[i][j]) {
          flag = 0;
          break;
        }
      }
      if (flag == 0) break;
    }
    if (flag) ans.push_back(t);
  }
  cout << ans.size() << endl;
  for (long long i = 0; i < ans.size(); i++) {
    cout << ans[i] << endl;
  }
  puts("");
  return 0;
}

### Prompt

Create a solution in cpp for 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 inf = 2147483647;
long long n;
long long bb[1111][11];
long long a[1111][11];
long long dian[1111][1111];
long long mo[1111];
long long cheng[1111][1111];
vector<long long> ans;
int main() {
  cin >> n;
  if (n > 100) {
    cout << 0 << endl;
    return 0;
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= 5; j++) cin >> bb[i][j];
  }
  for (long long t = 1; t <= n; t++) {
    memset(a, 0, sizeof(a));
    memset(dian, 0, sizeof(dian));
    memset(mo, 0, sizeof(mo));
    memset(cheng, 0, sizeof(cheng));
    long long num = 0;
    for (long long i = 1; i <= n; i++) {
      if (i == t) continue;
      num++;
      for (long long j = 1; j <= 5; j++) {
        a[num][j] = bb[i][j] - bb[t][j];
      }
    }
    for (long long i = 1; i <= num; i++) {
      for (long long j = i + 1; j <= num; j++) {
        for (long long k = 1; k <= 5; k++) {
          dian[i][j] += a[i][k] * a[j][k];
        }
        dian[j][i] = dian[i][j];
      }
    }
    for (long long i = 1; i <= num; i++) {
      for (long long j = 1; j <= 5; j++) {
        mo[i] += a[i][j] * a[i][j];
      }
    }
    for (long long i = 1; i <= num; i++) {
      for (long long j = i + 1; j <= num; j++) {
        cheng[i][j] = mo[i] * mo[j];
        cheng[j][i] = cheng[i][j];
      }
    }
    bool flag = 1;
    for (long long i = 1; i <= num; i++) {
      for (long long j = 1; j <= num; j++) {
        if (i == j) continue;
        if (dian[i][j] > 0 && dian[i][j] * dian[i][j] <= cheng[i][j]) {
          flag = 0;
          break;
        }
      }
      if (flag == 0) break;
    }
    if (flag) ans.push_back(t);
  }
  cout << ans.size() << endl;
  for (long long i = 0; i < ans.size(); i++) {
    cout << ans[i] << endl;
  }
  puts("");
  return 0;
}

```