#include <bits/stdc++.h>
const int maxn = 1000 + 10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
using namespace std;
struct Node {
  int a;
  int b;
  int c;
  int d;
  int e;
} T[maxn];
int ans[maxn], s = 0;
double dd(int i, int j) {
  return ((T[j].a - T[i].a) * (T[j].a - T[i].a) +
          (T[j].b - T[i].b) * (T[j].b - T[i].b) +
          (T[j].c - T[i].c) * (T[j].c - T[i].c) +
          (T[j].d - T[i].d) * (T[j].d - T[i].d) +
          (T[j].e - T[i].e) * (T[j].e - T[i].e));
}
double dx(int i, int j, int k) {
  return ((T[j].a - T[i].a) * (T[k].a - T[i].a) +
          (T[j].b - T[i].b) * (T[k].b - T[i].b) +
          (T[j].c - T[i].c) * (T[k].c - T[i].c) +
          (T[j].d - T[i].d) * (T[k].d - T[i].d) +
          (T[j].e - T[i].e) * (T[k].e - T[i].e));
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    scanf("%d%d%d%d%d", &T[i].a, &T[i].b, &T[i].c, &T[i].d, &T[i].e);
  if (n > 11) {
    cout << "0" << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    bool flag = false;
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      for (int k = 0; k < n; k++) {
        if (k == j || k == i) continue;
        double t = dx(i, j, k) / sqrt(dd(i, j) * dd(i, k));
        if (t <= 1 && t > 0) flag = true;
        if (flag) break;
      }
      if (flag) break;
    }
    if (flag) continue;
    ans[s++] = i + 1;
  }
  sort(ans, ans + s);
  cout << s << endl;
  for (int i = 0; i < s; i++) printf("%d\n", ans[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>
const int maxn = 1000 + 10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
using namespace std;
struct Node {
  int a;
  int b;
  int c;
  int d;
  int e;
} T[maxn];
int ans[maxn], s = 0;
double dd(int i, int j) {
  return ((T[j].a - T[i].a) * (T[j].a - T[i].a) +
          (T[j].b - T[i].b) * (T[j].b - T[i].b) +
          (T[j].c - T[i].c) * (T[j].c - T[i].c) +
          (T[j].d - T[i].d) * (T[j].d - T[i].d) +
          (T[j].e - T[i].e) * (T[j].e - T[i].e));
}
double dx(int i, int j, int k) {
  return ((T[j].a - T[i].a) * (T[k].a - T[i].a) +
          (T[j].b - T[i].b) * (T[k].b - T[i].b) +
          (T[j].c - T[i].c) * (T[k].c - T[i].c) +
          (T[j].d - T[i].d) * (T[k].d - T[i].d) +
          (T[j].e - T[i].e) * (T[k].e - T[i].e));
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    scanf("%d%d%d%d%d", &T[i].a, &T[i].b, &T[i].c, &T[i].d, &T[i].e);
  if (n > 11) {
    cout << "0" << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    bool flag = false;
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      for (int k = 0; k < n; k++) {
        if (k == j || k == i) continue;
        double t = dx(i, j, k) / sqrt(dd(i, j) * dd(i, k));
        if (t <= 1 && t > 0) flag = true;
        if (flag) break;
      }
      if (flag) break;
    }
    if (flag) continue;
    ans[s++] = i + 1;
  }
  sort(ans, ans + s);
  cout << s << endl;
  for (int i = 0; i < s; i++) printf("%d\n", ans[i]);
  return 0;
}

```