#include <bits/stdc++.h>
using namespace std;
struct s {
  int x, y[5];
} st;
bool operator<(const s &a, const s &b) { return a.x < a.x; }
int n, x[1005][5], ta;
vector<s> v;
vector<int> ans;
bool o;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", x[i] + j);
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) {
      st.x = 0;
      for (int k = 0; k < 5; k++) {
        st.x += ((x[i][k] - x[j][k]) * (x[i][k] - x[j][k]));
        st.y[k] = x[i][k] + x[j][k];
      }
      v.push_back(st);
    }
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    o = 1;
    for (int j = 0; j < v.size(); j++) {
      ta = 0;
      for (int k = 0; k < 5; k++)
        ta += ((v[j].y[k] - x[i][k] * 2) * (v[j].y[k] - x[i][k] * 2));
      if (ta > v[j].x) {
        o = 0;
        break;
      }
    }
    if (o) ans.push_back(i);
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) printf("%d\n", ans[i] + 1);
  return 0;
}

### Prompt

Your challenge is to write 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;
struct s {
  int x, y[5];
} st;
bool operator<(const s &a, const s &b) { return a.x < a.x; }
int n, x[1005][5], ta;
vector<s> v;
vector<int> ans;
bool o;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", x[i] + j);
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) {
      st.x = 0;
      for (int k = 0; k < 5; k++) {
        st.x += ((x[i][k] - x[j][k]) * (x[i][k] - x[j][k]));
        st.y[k] = x[i][k] + x[j][k];
      }
      v.push_back(st);
    }
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    o = 1;
    for (int j = 0; j < v.size(); j++) {
      ta = 0;
      for (int k = 0; k < 5; k++)
        ta += ((v[j].y[k] - x[i][k] * 2) * (v[j].y[k] - x[i][k] * 2));
      if (ta > v[j].x) {
        o = 0;
        break;
      }
    }
    if (o) ans.push_back(i);
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) printf("%d\n", ans[i] + 1);
  return 0;
}

```