#include <bits/stdc++.h>
using namespace std;
const int maxn = int(1e3) + 7;
int node[maxn][5], ans[maxn], cnt;
bool check(int i, int j, int k) {
  int ans = 0;
  for (int x = 0; x < 5; x++)
    ans += (node[i][x] - node[j][x]) * (node[k][x] - node[j][x]);
  return ans > 0;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", node[i] + j);
  if (n > 11) {
    return puts("0"), 0;
  }
  for (int i = 0; i < n; i++) {
    bool flag = true;
    for (int j = 0; j < n && flag; j++) {
      if (j == i) continue;
      for (int k = 0; k < n; k++) {
        if (k == j) continue;
        if (check(j, i, k)) {
          flag = false;
          break;
        }
      }
    }
    if (flag) ans[cnt++] = i + 1;
  }
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Generate 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 maxn = int(1e3) + 7;
int node[maxn][5], ans[maxn], cnt;
bool check(int i, int j, int k) {
  int ans = 0;
  for (int x = 0; x < 5; x++)
    ans += (node[i][x] - node[j][x]) * (node[k][x] - node[j][x]);
  return ans > 0;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", node[i] + j);
  if (n > 11) {
    return puts("0"), 0;
  }
  for (int i = 0; i < n; i++) {
    bool flag = true;
    for (int j = 0; j < n && flag; j++) {
      if (j == i) continue;
      for (int k = 0; k < n; k++) {
        if (k == j) continue;
        if (check(j, i, k)) {
          flag = false;
          break;
        }
      }
    }
    if (flag) ans[cnt++] = i + 1;
  }
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) printf("%d\n", ans[i]);
  return 0;
}

```