#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-8;
const long long MOD = 1e6 + 3;
const int INF = 0x3f3f3f3f;
const int maxn = 1e3 + 10;
const int maxm = 1e7 + 10;
struct INFO {
  int a, b, c, d, e;
} x[maxn], v1, v2;
int solve(INFO a, INFO b) {
  return a.a * b.a + a.b * b.b + a.c * b.c + a.d * b.d + a.e * b.e;
}
vector<int> ans;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i)
    scanf("%d%d%d%d%d", &x[i].a, &x[i].b, &x[i].c, &x[i].d, &x[i].e);
  int cnt = 0;
  for (int i = 1; i <= n; ++i) {
    bool flag = true;
    for (int j = 1; j <= n && flag; ++j)
      for (int k = 1; k <= n && flag; ++k) {
        if (i == j || i == k || j == k) continue;
        v1 = {x[i].a - x[j].a, x[i].b - x[j].b, x[i].c - x[j].c,
              x[i].d - x[j].d, x[i].e - x[j].e};
        v2 = {x[i].a - x[k].a, x[i].b - x[k].b, x[i].c - x[k].c,
              x[i].d - x[k].d, x[i].e - x[k].e};
        if (fabs(acos(1.0 * solve(v1, v2) /
                      (1.0 * solve(v1, v1) * solve(v2, v2)))) < pi / 2.0 - eps)
          flag = false;
      }
    if (flag) ans.push_back(i);
  }
  cout << ans.size() << '\n';
  if (!ans.empty()) {
    for (auto v : ans) printf("%d\n", v);
  }
  return 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;
const double pi = acos(-1.0);
const double eps = 1e-8;
const long long MOD = 1e6 + 3;
const int INF = 0x3f3f3f3f;
const int maxn = 1e3 + 10;
const int maxm = 1e7 + 10;
struct INFO {
  int a, b, c, d, e;
} x[maxn], v1, v2;
int solve(INFO a, INFO b) {
  return a.a * b.a + a.b * b.b + a.c * b.c + a.d * b.d + a.e * b.e;
}
vector<int> ans;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i)
    scanf("%d%d%d%d%d", &x[i].a, &x[i].b, &x[i].c, &x[i].d, &x[i].e);
  int cnt = 0;
  for (int i = 1; i <= n; ++i) {
    bool flag = true;
    for (int j = 1; j <= n && flag; ++j)
      for (int k = 1; k <= n && flag; ++k) {
        if (i == j || i == k || j == k) continue;
        v1 = {x[i].a - x[j].a, x[i].b - x[j].b, x[i].c - x[j].c,
              x[i].d - x[j].d, x[i].e - x[j].e};
        v2 = {x[i].a - x[k].a, x[i].b - x[k].b, x[i].c - x[k].c,
              x[i].d - x[k].d, x[i].e - x[k].e};
        if (fabs(acos(1.0 * solve(v1, v2) /
                      (1.0 * solve(v1, v1) * solve(v2, v2)))) < pi / 2.0 - eps)
          flag = false;
      }
    if (flag) ans.push_back(i);
  }
  cout << ans.size() << '\n';
  if (!ans.empty()) {
    for (auto v : ans) printf("%d\n", v);
  }
  return 0;
}

```