#include <bits/stdc++.h>
using namespace std;
const int N = 1100;
int n, ans[N];
struct point {
  long long x[5];
  int tag, i;
  void read(int id) {
    tag = -1;
    i = id;
    for (int i = 0; i < 5; i++) scanf("%lld", x + i);
  }
  long long operator*(point a) {
    long long res = 0;
    for (int i = 0; i < 5; i++) res += a.x[i] * x[i];
    return res;
  }
  point operator-(point a) {
    point tmp;
    for (int i = 0; i < 5; i++) tmp.x[i] = x[i] - a.x[i];
    return tmp;
  }
};
deque<point> a, b;
void chk(point &a, point &b, point &c) {
  point x = b - a, y = c - a;
  if (x * y > 0) {
    a.tag = 0;
    ans[a.i] = 0;
  }
}
void tri(point &a, point &b, point &c) {
  chk(a, b, c);
  chk(b, a, c);
  chk(c, a, b);
}
void overchk(point &a) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (i == j || i == a.i || j == a.i) continue;
      chk(a, b[i], b[j]);
      if (a.tag == 0) return;
    }
  a.tag = 1;
  ans[a.i] = 1;
}
int main() {
  memset(ans, -1, sizeof(ans));
  cin >> n;
  a.resize(n);
  b.resize(n);
  for (int i = 0; i < n; i++) a[i].read(i + 1), b[i] = a[i];
  while (a.size() >= 3) {
    point ap = a[0], bp = a[1], cp = a[2];
    for (int i = 0; i < 3; i++) a.pop_front();
    tri(ap, bp, cp);
    if (ap.tag < 0) a.push_back(ap);
    if (bp.tag < 0) a.push_back(bp);
    if (cp.tag < 0) a.push_back(cp);
  }
  for (int i = 0; i < a.size(); i++) overchk(a[i]);
  int tot = 0;
  for (int i = 1; i <= n; i++)
    if (ans[i]) tot++;
  cout << tot << endl;
  for (int i = 1; i <= n; i++)
    if (ans[i]) printf("%d\n", i);
  return 0;
}

### Prompt

Generate 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 N = 1100;
int n, ans[N];
struct point {
  long long x[5];
  int tag, i;
  void read(int id) {
    tag = -1;
    i = id;
    for (int i = 0; i < 5; i++) scanf("%lld", x + i);
  }
  long long operator*(point a) {
    long long res = 0;
    for (int i = 0; i < 5; i++) res += a.x[i] * x[i];
    return res;
  }
  point operator-(point a) {
    point tmp;
    for (int i = 0; i < 5; i++) tmp.x[i] = x[i] - a.x[i];
    return tmp;
  }
};
deque<point> a, b;
void chk(point &a, point &b, point &c) {
  point x = b - a, y = c - a;
  if (x * y > 0) {
    a.tag = 0;
    ans[a.i] = 0;
  }
}
void tri(point &a, point &b, point &c) {
  chk(a, b, c);
  chk(b, a, c);
  chk(c, a, b);
}
void overchk(point &a) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (i == j || i == a.i || j == a.i) continue;
      chk(a, b[i], b[j]);
      if (a.tag == 0) return;
    }
  a.tag = 1;
  ans[a.i] = 1;
}
int main() {
  memset(ans, -1, sizeof(ans));
  cin >> n;
  a.resize(n);
  b.resize(n);
  for (int i = 0; i < n; i++) a[i].read(i + 1), b[i] = a[i];
  while (a.size() >= 3) {
    point ap = a[0], bp = a[1], cp = a[2];
    for (int i = 0; i < 3; i++) a.pop_front();
    tri(ap, bp, cp);
    if (ap.tag < 0) a.push_back(ap);
    if (bp.tag < 0) a.push_back(bp);
    if (cp.tag < 0) a.push_back(cp);
  }
  for (int i = 0; i < a.size(); i++) overchk(a[i]);
  int tot = 0;
  for (int i = 1; i <= n; i++)
    if (ans[i]) tot++;
  cout << tot << endl;
  for (int i = 1; i <= n; i++)
    if (ans[i]) printf("%d\n", i);
  return 0;
}

```