#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 7;
const int INF = 1e9 + 7;
int a[maxn][7];
int vis[maxn];
bool check(int x, int y, int z) {
  int i, j;
  int xx[6], yy[6];
  for (i = 1; i <= 5; i++) {
    xx[i] = a[y][i] - a[x][i];
    yy[i] = a[z][i] - a[x][i];
  }
  double ans = 0;
  for (i = 1; i <= 5; i++) {
    ans = ans + xx[i] * yy[i];
  }
  if (ans <= 0)
    return true;
  else
    return false;
}
int main() {
  int i, j, n, m, t, z;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= 5; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  int k;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (j == i) continue;
      for (k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        if (!check(i, j, k)) break;
      }
      if (k <= n) break;
    }
    if (j > n) {
      vis[i] = 1;
    }
  }
  vector<int> ans;
  for (i = 1; i <= n; i++) {
    if (vis[i]) ans.push_back(i);
  }
  printf("%d\n", ans.size());
  for (i = 0; i < ans.size(); i++) {
    printf("%d%c", ans[i], i == ans.size() ? '\n' : ' ');
  }
  return 0;
}

### Prompt

Please formulate 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 = 1e4 + 7;
const int INF = 1e9 + 7;
int a[maxn][7];
int vis[maxn];
bool check(int x, int y, int z) {
  int i, j;
  int xx[6], yy[6];
  for (i = 1; i <= 5; i++) {
    xx[i] = a[y][i] - a[x][i];
    yy[i] = a[z][i] - a[x][i];
  }
  double ans = 0;
  for (i = 1; i <= 5; i++) {
    ans = ans + xx[i] * yy[i];
  }
  if (ans <= 0)
    return true;
  else
    return false;
}
int main() {
  int i, j, n, m, t, z;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= 5; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  int k;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (j == i) continue;
      for (k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        if (!check(i, j, k)) break;
      }
      if (k <= n) break;
    }
    if (j > n) {
      vis[i] = 1;
    }
  }
  vector<int> ans;
  for (i = 1; i <= n; i++) {
    if (vis[i]) ans.push_back(i);
  }
  printf("%d\n", ans.size());
  for (i = 0; i < ans.size(); i++) {
    printf("%d%c", ans[i], i == ans.size() ? '\n' : ' ');
  }
  return 0;
}

```