#include <bits/stdc++.h>
using namespace std;
int x[1013][5];
bool chk(int a, int b, int c) {
  int s = 0;
  for (int i = int(0), __border_righti = int(5); i < __border_righti; i++) {
    s += (x[b][i] - x[a][i]) * (x[c][i] - x[a][i]);
  }
  return s <= 0;
}
int main() {
  int N = 0;
  scanf("%d", &N);
  for (int i = int(0), __border_righti = int(N); i < __border_righti; i++) {
    for (int j = int(0), __border_rightj = int(5); j < __border_rightj; j++) {
      scanf("%d", &x[i][j]);
    }
  }
  if (N > 32) {
    printf("%d\n", 0);
  } else {
    int ans[40];
    int n = 0;
    for (int i = int(0), __border_righti = int(N); i < __border_righti; i++) {
      bool flag = true;
      for (int j = int(0), __border_rightj = int(N); j < __border_rightj; j++) {
        if (j == i) {
          continue;
        }
        for (int k = int(0), __border_rightk = int(j); k < __border_rightk;
             k++) {
          if (k == i) {
            continue;
          }
          if (!chk(i, j, k)) {
            flag = false;
          }
        }
      }
      if (flag) {
        ans[n++] = i + 1;
      }
    }
    printf("%d\n", n);
    for (int i = int(0), __border_righti = int(n); i < __border_righti; i++) {
      printf("%d\n", ans[i]);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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 x[1013][5];
bool chk(int a, int b, int c) {
  int s = 0;
  for (int i = int(0), __border_righti = int(5); i < __border_righti; i++) {
    s += (x[b][i] - x[a][i]) * (x[c][i] - x[a][i]);
  }
  return s <= 0;
}
int main() {
  int N = 0;
  scanf("%d", &N);
  for (int i = int(0), __border_righti = int(N); i < __border_righti; i++) {
    for (int j = int(0), __border_rightj = int(5); j < __border_rightj; j++) {
      scanf("%d", &x[i][j]);
    }
  }
  if (N > 32) {
    printf("%d\n", 0);
  } else {
    int ans[40];
    int n = 0;
    for (int i = int(0), __border_righti = int(N); i < __border_righti; i++) {
      bool flag = true;
      for (int j = int(0), __border_rightj = int(N); j < __border_rightj; j++) {
        if (j == i) {
          continue;
        }
        for (int k = int(0), __border_rightk = int(j); k < __border_rightk;
             k++) {
          if (k == i) {
            continue;
          }
          if (!chk(i, j, k)) {
            flag = false;
          }
        }
      }
      if (flag) {
        ans[n++] = i + 1;
      }
    }
    printf("%d\n", n);
    for (int i = int(0), __border_righti = int(n); i < __border_righti; i++) {
      printf("%d\n", ans[i]);
    }
  }
  return 0;
}

```