#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n, a[N][7];
vector<int> idx;
int dot(int o, int i, int j) {
  int res = 0;
  for (int k = 0; k < 5; ++k) {
    res += (a[i][k] - a[o][k]) * (a[j][k] - a[o][k]);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < 5; ++j) {
      scanf("%d", a[i] + j);
    }
  }
  for (int i = 1; i <= n; ++i) {
    int good = 1;
    for (int j = 1; j <= n; ++j) {
      if (j == i) continue;
      for (int k = 1; k <= n; ++k) {
        if (k == i or k == j) continue;
        if (dot(i, j, k) > 0) {
          good = 0;
          break;
        }
      }
      if (good == 0) break;
    }
    if (good) {
      idx.push_back(i);
    }
  }
  printf("%d\n", (int)idx.size());
  for (int x : idx) printf("%d\n", x);
  return 0;
}

### Prompt

Please create a solution in Cpp 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, a[N][7];
vector<int> idx;
int dot(int o, int i, int j) {
  int res = 0;
  for (int k = 0; k < 5; ++k) {
    res += (a[i][k] - a[o][k]) * (a[j][k] - a[o][k]);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < 5; ++j) {
      scanf("%d", a[i] + j);
    }
  }
  for (int i = 1; i <= n; ++i) {
    int good = 1;
    for (int j = 1; j <= n; ++j) {
      if (j == i) continue;
      for (int k = 1; k <= n; ++k) {
        if (k == i or k == j) continue;
        if (dot(i, j, k) > 0) {
          good = 0;
          break;
        }
      }
      if (good == 0) break;
    }
    if (good) {
      idx.push_back(i);
    }
  }
  printf("%d\n", (int)idx.size());
  for (int x : idx) printf("%d\n", x);
  return 0;
}

```