#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000 + 5;
const double pi = acos(-1);
int n;
struct node {
  int a, b, c, d, e;
  double getlen() { return sqrt(a * a + b * b + c * c + d * d + e * e); }
  double operator*(const node &temp) const {
    return a * 1.0 * temp.a + b * temp.b + c * temp.c + d * temp.d + e * temp.e;
  }
  node operator-(const node &temp) const {
    node ans;
    ans.a = a - temp.a;
    ans.b = b - temp.b;
    ans.c = c - temp.c;
    ans.d = d - temp.d;
    ans.e = e - temp.e;
    return ans;
  }
} p[MAXN];
bool check(node a, node b, node c) {
  node ab = b - a;
  node ac = c - a;
  if (acos(ab * ac / (ab.getlen() * ac.getlen())) < pi * 0.5) return 1;
  return 0;
}
vector<int> ans;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i)
    scanf("%d%d%d%d%d", &p[i].a, &p[i].b, &p[i].c, &p[i].d, &p[i].e);
  for (int i = 0; i < n; ++i) {
    bool flag = 1;
    for (int j = 0; j < n; ++j)
      if (j != i && flag)
        for (int k = 0; k < n; ++k)
          if (k != i && k != j && flag) {
            if (check(p[i], p[j], p[k])) {
              flag = 0;
              break;
            }
          }
    if (flag) ans.push_back(i + 1);
  }
  int siz = ans.size();
  printf("%d\n", siz);
  for (int i = 0; i < siz; ++i) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Create a solution in cpp for 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 = 1000 + 5;
const double pi = acos(-1);
int n;
struct node {
  int a, b, c, d, e;
  double getlen() { return sqrt(a * a + b * b + c * c + d * d + e * e); }
  double operator*(const node &temp) const {
    return a * 1.0 * temp.a + b * temp.b + c * temp.c + d * temp.d + e * temp.e;
  }
  node operator-(const node &temp) const {
    node ans;
    ans.a = a - temp.a;
    ans.b = b - temp.b;
    ans.c = c - temp.c;
    ans.d = d - temp.d;
    ans.e = e - temp.e;
    return ans;
  }
} p[MAXN];
bool check(node a, node b, node c) {
  node ab = b - a;
  node ac = c - a;
  if (acos(ab * ac / (ab.getlen() * ac.getlen())) < pi * 0.5) return 1;
  return 0;
}
vector<int> ans;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i)
    scanf("%d%d%d%d%d", &p[i].a, &p[i].b, &p[i].c, &p[i].d, &p[i].e);
  for (int i = 0; i < n; ++i) {
    bool flag = 1;
    for (int j = 0; j < n; ++j)
      if (j != i && flag)
        for (int k = 0; k < n; ++k)
          if (k != i && k != j && flag) {
            if (check(p[i], p[j], p[k])) {
              flag = 0;
              break;
            }
          }
    if (flag) ans.push_back(i + 1);
  }
  int siz = ans.size();
  printf("%d\n", siz);
  for (int i = 0; i < siz; ++i) printf("%d\n", ans[i]);
  return 0;
}

```