#include <bits/stdc++.h>
const double pi = acos(-1.);
const int max_N = 1e3;
std::unordered_set<long long> set;
struct poi {
  int d[5];
  inline void init() {
    for (int i = 0; i < 5; ++i) scanf("%d", d + i);
  }
} p[max_N + 21];
int n;
bool ban[max_N + 21];
inline bool check(int a, int b, int c) {
  poi ab, ac;
  for (int i = 0; i < 5; ++i) {
    ab.d[i] = p[b].d[i] - p[a].d[i];
    ac.d[i] = p[c].d[i] - p[a].d[i];
  }
  double abac = 0, AB = 0, AC = 0;
  for (int i = 0; i < 5; ++i) {
    abac += ab.d[i] * ac.d[i];
    AB += ab.d[i] * ab.d[i];
    AC += ac.d[i] * ac.d[i];
  }
  AB = std::sqrt(AB);
  AC = std::sqrt(AC);
  double angle = acos(abac / AB / AC);
  return angle * 2 < pi;
}
std::vector<int> ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) p[i].init();
  if (n > 100) return puts("0"), 0;
  for (int a = 1; a <= n; ++a)
    for (int b = 1; b <= n; ++b)
      if (a != b)
        for (int c = 1; c <= n; ++c)
          if (a != c) {
            if (b == c) continue;
            if (check(a, b, c)) ban[a] = 1;
          }
  for (int i = 1; i <= n; ++i)
    if (!ban[i]) ans.push_back(i);
  int ans1 = ans.size();
  printf("%d\n", ans1);
  for (int i = 0; i < ans1; ++i) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
const double pi = acos(-1.);
const int max_N = 1e3;
std::unordered_set<long long> set;
struct poi {
  int d[5];
  inline void init() {
    for (int i = 0; i < 5; ++i) scanf("%d", d + i);
  }
} p[max_N + 21];
int n;
bool ban[max_N + 21];
inline bool check(int a, int b, int c) {
  poi ab, ac;
  for (int i = 0; i < 5; ++i) {
    ab.d[i] = p[b].d[i] - p[a].d[i];
    ac.d[i] = p[c].d[i] - p[a].d[i];
  }
  double abac = 0, AB = 0, AC = 0;
  for (int i = 0; i < 5; ++i) {
    abac += ab.d[i] * ac.d[i];
    AB += ab.d[i] * ab.d[i];
    AC += ac.d[i] * ac.d[i];
  }
  AB = std::sqrt(AB);
  AC = std::sqrt(AC);
  double angle = acos(abac / AB / AC);
  return angle * 2 < pi;
}
std::vector<int> ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) p[i].init();
  if (n > 100) return puts("0"), 0;
  for (int a = 1; a <= n; ++a)
    for (int b = 1; b <= n; ++b)
      if (a != b)
        for (int c = 1; c <= n; ++c)
          if (a != c) {
            if (b == c) continue;
            if (check(a, b, c)) ban[a] = 1;
          }
  for (int i = 1; i <= n; ++i)
    if (!ban[i]) ans.push_back(i);
  int ans1 = ans.size();
  printf("%d\n", ans1);
  for (int i = 0; i < ans1; ++i) printf("%d\n", ans[i]);
  return 0;
}

```