#include <bits/stdc++.h>
using namespace std;
inline int Read() {
  int e = 0;
  char ch = getchar();
  while (ch < '0' || '9' < ch) ch = getchar();
  while ('0' <= ch && ch <= '9') e = e * 10 + (ch - '0'), ch = getchar();
  return e;
}
const int maxn = 1e3 + 7;
struct node {
  int a, b, c, d, e;
  void Scanf() { cin >> a >> b >> c >> d >> e; }
  node operator-(const node& T) const {
    node tmp;
    tmp.a = a - T.a;
    tmp.b = b - T.b;
    tmp.c = c - T.c;
    tmp.d = d - T.d;
    tmp.e = e - T.e;
    return tmp;
  }
  int operator*(const node& T) const {
    return a * T.a + b * T.b + c * T.c + d * T.d + e * T.e;
  }
  int ABS() { return a * a + b * b + c * c + d * d + e * e; }
} a[maxn];
bool vis[maxn];
bool Judge(int k, int u, int v) {
  node tmp1 = a[u] - a[k], tmp2 = a[v] - a[k];
  int k1 = tmp1 * tmp2;
  return k1 <= 0;
}
int main() {
  int n;
  cin >> n;
  int sum = n;
  for (int i = 1; i <= n; i++) a[i].Scanf();
  for (int i = 1; i < n; i++) {
    for (int j = i + 1; j <= n; j++) {
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j || vis[k]) continue;
        if (Judge(k, i, j)) continue;
        vis[k] = true;
        sum--;
      }
    }
  }
  cout << sum << endl;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) cout << i << endl;
  return 0;
}

### 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;
inline int Read() {
  int e = 0;
  char ch = getchar();
  while (ch < '0' || '9' < ch) ch = getchar();
  while ('0' <= ch && ch <= '9') e = e * 10 + (ch - '0'), ch = getchar();
  return e;
}
const int maxn = 1e3 + 7;
struct node {
  int a, b, c, d, e;
  void Scanf() { cin >> a >> b >> c >> d >> e; }
  node operator-(const node& T) const {
    node tmp;
    tmp.a = a - T.a;
    tmp.b = b - T.b;
    tmp.c = c - T.c;
    tmp.d = d - T.d;
    tmp.e = e - T.e;
    return tmp;
  }
  int operator*(const node& T) const {
    return a * T.a + b * T.b + c * T.c + d * T.d + e * T.e;
  }
  int ABS() { return a * a + b * b + c * c + d * d + e * e; }
} a[maxn];
bool vis[maxn];
bool Judge(int k, int u, int v) {
  node tmp1 = a[u] - a[k], tmp2 = a[v] - a[k];
  int k1 = tmp1 * tmp2;
  return k1 <= 0;
}
int main() {
  int n;
  cin >> n;
  int sum = n;
  for (int i = 1; i <= n; i++) a[i].Scanf();
  for (int i = 1; i < n; i++) {
    for (int j = i + 1; j <= n; j++) {
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j || vis[k]) continue;
        if (Judge(k, i, j)) continue;
        vis[k] = true;
        sum--;
      }
    }
  }
  cout << sum << endl;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) cout << i << endl;
  return 0;
}

```