#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n, x[N][5], y[N][5];
bool ok[N];
int main() {
  scanf("%d", &n);
  if (n >= 100) return puts("0"), 0;
  for (int i = 1; i <= n; i++) ok[i] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", &x[i][j]);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++)
      for (int k = 0; k < 5; k++) y[j][k] = x[j][k] - x[i][k];
    for (int j = 1; j <= n; j++)
      for (int k = 1; k <= n; k++)
        if (j != i && k != i && j != k) {
          int mul = 0;
          for (int a = 0; a < 5; a++) mul += y[j][a] * y[k][a];
          if (mul > 0) ok[i] = 0;
        }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    if (ok[i]) ans++;
  printf("%d\n", ans);
  for (int i = 1; i <= n; i++)
    if (ok[i]) printf("%d\n", 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 N = 1010;
int n, x[N][5], y[N][5];
bool ok[N];
int main() {
  scanf("%d", &n);
  if (n >= 100) return puts("0"), 0;
  for (int i = 1; i <= n; i++) ok[i] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", &x[i][j]);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++)
      for (int k = 0; k < 5; k++) y[j][k] = x[j][k] - x[i][k];
    for (int j = 1; j <= n; j++)
      for (int k = 1; k <= n; k++)
        if (j != i && k != i && j != k) {
          int mul = 0;
          for (int a = 0; a < 5; a++) mul += y[j][a] * y[k][a];
          if (mul > 0) ok[i] = 0;
        }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    if (ok[i]) ans++;
  printf("%d\n", ans);
  for (int i = 1; i <= n; i++)
    if (ok[i]) printf("%d\n", i);
  return 0;
}

```