#include <bits/stdc++.h>
using namespace std;
inline int rd() {
  int x = 0, c = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar()) f = c != '-';
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  return f ? x : -x;
}
const int CH = 5, N = 111;
int n;
struct node {
  int a[CH];
  node operator-(const node& t) const {
    node temp;
    for (int i = (0), i_end = (CH - 1); i <= i_end; i++)
      temp.a[i] = a[i] - t.a[i];
    return temp;
  }
  long long operator*(const node& t) const {
    long long ret = 0;
    for (int i = (0), i_end = (CH - 1); i <= i_end; i++)
      ret += 1LL * a[i] * t.a[i];
    return ret;
  }
} p[N];
bool judge(int x) {
  for (int i = (1), i_end = (n); i <= i_end; i++)
    if (i != x) {
      for (int j = (1), j_end = (n); j <= j_end; j++)
        if (j != x && j != i) {
          node s1 = p[x] - p[i], s2 = p[x] - p[j];
          long long tt = s1 * s2;
          if (tt > 0) return 0;
        }
    }
  return 1;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  if (n > 100) {
    puts("0");
    return 0;
  }
  for (int i = (1), i_end = (n); i <= i_end; i++) {
    for (int j = (0), j_end = (CH - 1); j <= j_end; j++) {
      cin >> p[i].a[j];
    }
  }
  if (n <= 2) {
    printf("%d\n", n);
    for (int i = (1), i_end = (n); i <= i_end; i++) printf("%d\n", i);
    return 0;
  }
  for (int i = (1), i_end = (n); i <= i_end; i++)
    if (judge(i)) {
      puts("1");
      printf("%d\n", i);
      return 0;
    }
  puts("0");
}

### Prompt

Your task is to create 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;
inline int rd() {
  int x = 0, c = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar()) f = c != '-';
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  return f ? x : -x;
}
const int CH = 5, N = 111;
int n;
struct node {
  int a[CH];
  node operator-(const node& t) const {
    node temp;
    for (int i = (0), i_end = (CH - 1); i <= i_end; i++)
      temp.a[i] = a[i] - t.a[i];
    return temp;
  }
  long long operator*(const node& t) const {
    long long ret = 0;
    for (int i = (0), i_end = (CH - 1); i <= i_end; i++)
      ret += 1LL * a[i] * t.a[i];
    return ret;
  }
} p[N];
bool judge(int x) {
  for (int i = (1), i_end = (n); i <= i_end; i++)
    if (i != x) {
      for (int j = (1), j_end = (n); j <= j_end; j++)
        if (j != x && j != i) {
          node s1 = p[x] - p[i], s2 = p[x] - p[j];
          long long tt = s1 * s2;
          if (tt > 0) return 0;
        }
    }
  return 1;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  if (n > 100) {
    puts("0");
    return 0;
  }
  for (int i = (1), i_end = (n); i <= i_end; i++) {
    for (int j = (0), j_end = (CH - 1); j <= j_end; j++) {
      cin >> p[i].a[j];
    }
  }
  if (n <= 2) {
    printf("%d\n", n);
    for (int i = (1), i_end = (n); i <= i_end; i++) printf("%d\n", i);
    return 0;
  }
  for (int i = (1), i_end = (n); i <= i_end; i++)
    if (judge(i)) {
      puts("1");
      printf("%d\n", i);
      return 0;
    }
  puts("0");
}

```