#include <bits/stdc++.h>
using namespace std;
int n, a[1000][5], ans[20];
int main() {
  scanf("%d", &n);
  if (n > 11) {
    printf("0");
    return 0;
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 5; ++j) scanf("%d", &a[i][j]);
  for (int i = 0; i < n; ++i) {
    int ok = 1;
    for (int j = 0; j < n; ++j)
      if (j != i)
        for (int k = 0; k < n; ++k)
          if (k != i && k != j) {
            int tmp = 0;
            for (int t = 0; t < 5; ++t)
              tmp += (a[j][t] - a[i][t]) * (a[k][t] - a[i][t]);
            if (tmp > 0) ok = 0;
          }
    if (!ok) continue;
    ans[++ans[0]] = i + 1;
  }
  for (int i = 0; i <= ans[0]; ++i) printf("%d\n", ans[i]);
}

### Prompt

Construct a cpp code solution to the problem outlined:

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[1000][5], ans[20];
int main() {
  scanf("%d", &n);
  if (n > 11) {
    printf("0");
    return 0;
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 5; ++j) scanf("%d", &a[i][j]);
  for (int i = 0; i < n; ++i) {
    int ok = 1;
    for (int j = 0; j < n; ++j)
      if (j != i)
        for (int k = 0; k < n; ++k)
          if (k != i && k != j) {
            int tmp = 0;
            for (int t = 0; t < 5; ++t)
              tmp += (a[j][t] - a[i][t]) * (a[k][t] - a[i][t]);
            if (tmp > 0) ok = 0;
          }
    if (!ok) continue;
    ans[++ans[0]] = i + 1;
  }
  for (int i = 0; i <= ans[0]; ++i) printf("%d\n", ans[i]);
}

```