#include <bits/stdc++.h>
using namespace std;
long long a[1024 + 5];
long long b[1024 + 5];
long long c[1024 + 5];
long long d[1024 + 5];
long long e[1024 + 5];
long long dist[1024 + 5][1024 + 5];
long long Get(int i, int j) {
  long long u = ((a[i] - a[j]) * (a[i] - a[j]));
  long long v = ((b[i] - b[j]) * (b[i] - b[j]));
  long long w = ((c[i] - c[j]) * (c[i] - c[j]));
  long long t = ((d[i] - d[j]) * (d[i] - d[j]));
  long long s = ((e[i] - e[j]) * (e[i] - e[j]));
  return u + v + w + t + s;
}
bool mark[1024 + 5];
int main() {
  memset((mark), (false), sizeof((mark)));
  int n;
  scanf("%d", &n);
  for (int(i) = 0; (i) < (n); (i)++) {
    scanf("%lld %lld %lld %lld %lld", &a[i], &b[i], &c[i], &d[i], &e[i]);
  }
  for (int(i) = 0; (i) < (n); (i)++) {
    for (int(j) = 0; (j) < (n); (j)++) {
      if (i == j) {
        continue;
      }
      dist[i][j] = Get(i, j);
    }
  }
  long long u, v, w;
  for (int(i) = 0; (i) < (n); (i)++) {
    for (int(j) = (i + 1); (j) < (n); (j)++) {
      for (int(r) = (j + 1); (r) < (n); (r)++) {
        u = dist[i][j];
        v = dist[j][r];
        w = dist[i][r];
        if (w < u + v) {
          mark[j] = 1;
        }
        if (u < v + w) {
          mark[r] = 1;
        }
        if (v < u + w) {
          mark[i] = 1;
        }
      }
    }
  }
  vector<int> p;
  int ans = 0;
  for (int(i) = 0; (i) < (n); (i)++) {
    if (!mark[i]) {
      ++ans;
      p.push_back(i + 1);
    }
  }
  printf("%d\n", ans);
  for (int(i) = 0; (i) < ((int)(p).size()); (i)++) {
    printf("%d\n", p[i]);
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
long long a[1024 + 5];
long long b[1024 + 5];
long long c[1024 + 5];
long long d[1024 + 5];
long long e[1024 + 5];
long long dist[1024 + 5][1024 + 5];
long long Get(int i, int j) {
  long long u = ((a[i] - a[j]) * (a[i] - a[j]));
  long long v = ((b[i] - b[j]) * (b[i] - b[j]));
  long long w = ((c[i] - c[j]) * (c[i] - c[j]));
  long long t = ((d[i] - d[j]) * (d[i] - d[j]));
  long long s = ((e[i] - e[j]) * (e[i] - e[j]));
  return u + v + w + t + s;
}
bool mark[1024 + 5];
int main() {
  memset((mark), (false), sizeof((mark)));
  int n;
  scanf("%d", &n);
  for (int(i) = 0; (i) < (n); (i)++) {
    scanf("%lld %lld %lld %lld %lld", &a[i], &b[i], &c[i], &d[i], &e[i]);
  }
  for (int(i) = 0; (i) < (n); (i)++) {
    for (int(j) = 0; (j) < (n); (j)++) {
      if (i == j) {
        continue;
      }
      dist[i][j] = Get(i, j);
    }
  }
  long long u, v, w;
  for (int(i) = 0; (i) < (n); (i)++) {
    for (int(j) = (i + 1); (j) < (n); (j)++) {
      for (int(r) = (j + 1); (r) < (n); (r)++) {
        u = dist[i][j];
        v = dist[j][r];
        w = dist[i][r];
        if (w < u + v) {
          mark[j] = 1;
        }
        if (u < v + w) {
          mark[r] = 1;
        }
        if (v < u + w) {
          mark[i] = 1;
        }
      }
    }
  }
  vector<int> p;
  int ans = 0;
  for (int(i) = 0; (i) < (n); (i)++) {
    if (!mark[i]) {
      ++ans;
      p.push_back(i + 1);
    }
  }
  printf("%d\n", ans);
  for (int(i) = 0; (i) < ((int)(p).size()); (i)++) {
    printf("%d\n", p[i]);
  }
  return 0;
}

```