#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
int p[SIZE][5];
int main() {
  int(n);
  scanf("%d", &n);
  for (int i = (1); i < (n + 1); ++i) {
    for (int j = 0; j < (5); ++j) scanf("%d", &(p[i][j]));
  }
  vector<int> an;
  for (int i = (1); i < (n + 1); ++i) {
    bool fail = 0;
    for (int j = (1); j < (n + 1); ++j) {
      if (j == i) continue;
      int d1[5];
      for (int r = 0; r < (5); ++r) d1[r] = p[j][r] - p[i][r];
      for (int k = (1); k < (j); ++k) {
        int d2[5] = {};
        int v = 0;
        for (int r = 0; r < (5); ++r) v += (p[k][r] - p[i][r]) * d1[r];
        if (v > 0) {
          fail = 1;
          break;
        }
      }
      if (fail) break;
    }
    if (!fail) an.push_back(i);
  }
  printf("%d\n", ((int)(an).size()));
  for (int i = 0; i < (((int)(an).size())); ++i)
    printf("%d%c", an[i], " \n"[i == ((int)(an).size()) - 1]);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
int p[SIZE][5];
int main() {
  int(n);
  scanf("%d", &n);
  for (int i = (1); i < (n + 1); ++i) {
    for (int j = 0; j < (5); ++j) scanf("%d", &(p[i][j]));
  }
  vector<int> an;
  for (int i = (1); i < (n + 1); ++i) {
    bool fail = 0;
    for (int j = (1); j < (n + 1); ++j) {
      if (j == i) continue;
      int d1[5];
      for (int r = 0; r < (5); ++r) d1[r] = p[j][r] - p[i][r];
      for (int k = (1); k < (j); ++k) {
        int d2[5] = {};
        int v = 0;
        for (int r = 0; r < (5); ++r) v += (p[k][r] - p[i][r]) * d1[r];
        if (v > 0) {
          fail = 1;
          break;
        }
      }
      if (fail) break;
    }
    if (!fail) an.push_back(i);
  }
  printf("%d\n", ((int)(an).size()));
  for (int i = 0; i < (((int)(an).size())); ++i)
    printf("%d%c", an[i], " \n"[i == ((int)(an).size()) - 1]);
  return 0;
}

```