#include <bits/stdc++.h>
using namespace std;
int n, a[1001000][6], bad[1001000], b, c, d;
queue<int> s;
vector<int> ans;
int scalar(int i, int j, int p) {
  return (a[i][0] - a[p][0]) * (a[j][0] - a[p][0]) +
         (a[i][1] - a[p][1]) * (a[j][1] - a[p][1]) +
         (a[i][2] - a[p][2]) * (a[j][2] - a[p][2]) +
         (a[i][3] - a[p][3]) * (a[j][3] - a[p][3]) +
         (a[i][4] - a[p][4]) * (a[j][4] - a[p][4]);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", &a[i][j]);
      s.push(i);
    }
  }
  while (!s.empty()) {
    if (s.size() > 2) {
      b = s.front();
      s.pop();
      c = s.front();
      s.pop();
      d = s.front();
      s.pop();
      if (scalar(d, b, c) <= 0) {
        bad[d] = 1;
        bad[b] = 1;
        s.push(c);
      } else if (scalar(d, c, b) <= 0) {
        bad[d] = 1;
        bad[c] = 1;
        s.push(b);
      } else if (scalar(b, c, d) <= 0) {
        bad[b] = 1;
        bad[c] = 1;
        s.push(d);
      } else {
        bad[d] = 1;
        bad[b] = 1;
        bad[c] = 1;
      }
    } else {
      b = s.front();
      s.pop();
      int flag = 1;
      for (int i = 0; i < n && flag; i++) {
        if (i == b) continue;
        for (int j = 0; j < n && flag; j++) {
          if (j == b || j == i) continue;
          if (scalar(i, j, b) > 0) flag = 0;
        }
      }
      if (flag)
        ans.push_back(b);
      else
        bad[b] = 1;
    }
  }
  sort(ans.begin(), ans.end());
  printf("%d\n", ans.size());
  for (int x : ans) printf("%d\n", x + 1);
  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;
int n, a[1001000][6], bad[1001000], b, c, d;
queue<int> s;
vector<int> ans;
int scalar(int i, int j, int p) {
  return (a[i][0] - a[p][0]) * (a[j][0] - a[p][0]) +
         (a[i][1] - a[p][1]) * (a[j][1] - a[p][1]) +
         (a[i][2] - a[p][2]) * (a[j][2] - a[p][2]) +
         (a[i][3] - a[p][3]) * (a[j][3] - a[p][3]) +
         (a[i][4] - a[p][4]) * (a[j][4] - a[p][4]);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", &a[i][j]);
      s.push(i);
    }
  }
  while (!s.empty()) {
    if (s.size() > 2) {
      b = s.front();
      s.pop();
      c = s.front();
      s.pop();
      d = s.front();
      s.pop();
      if (scalar(d, b, c) <= 0) {
        bad[d] = 1;
        bad[b] = 1;
        s.push(c);
      } else if (scalar(d, c, b) <= 0) {
        bad[d] = 1;
        bad[c] = 1;
        s.push(b);
      } else if (scalar(b, c, d) <= 0) {
        bad[b] = 1;
        bad[c] = 1;
        s.push(d);
      } else {
        bad[d] = 1;
        bad[b] = 1;
        bad[c] = 1;
      }
    } else {
      b = s.front();
      s.pop();
      int flag = 1;
      for (int i = 0; i < n && flag; i++) {
        if (i == b) continue;
        for (int j = 0; j < n && flag; j++) {
          if (j == b || j == i) continue;
          if (scalar(i, j, b) > 0) flag = 0;
        }
      }
      if (flag)
        ans.push_back(b);
      else
        bad[b] = 1;
    }
  }
  sort(ans.begin(), ans.end());
  printf("%d\n", ans.size());
  for (int x : ans) printf("%d\n", x + 1);
  return 0;
}

```