#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int maxn = 1e5 + 10;
struct node {
  int a[5];
} p[maxn];
vector<int> ans;
int calc(int x, int y, int z) {
  int res = 0;
  for (int i = 0; i < 5; i++) {
    res += (p[y].a[i] - p[x].a[i]) * (p[z].a[i] - p[x].a[i]);
  }
  return res;
}
int main() {
  int n = read();
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", &p[i].a[j]);
  for (int i = 1; i <= n; i++) {
    bool f = 0;
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      for (int k = j + 1; k <= n; k++) {
        if (k == i) continue;
        if (calc(i, j, k) > 0) {
          f = 1;
          break;
        }
      }
      if (f) break;
    }
    if (!f) ans.push_back(i);
  }
  cout << ans.size() << endl;
  for (int i = 0; i < (int)ans.size(); i++) cout << ans[i] << " ";
  if (ans.size()) puts("");
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int maxn = 1e5 + 10;
struct node {
  int a[5];
} p[maxn];
vector<int> ans;
int calc(int x, int y, int z) {
  int res = 0;
  for (int i = 0; i < 5; i++) {
    res += (p[y].a[i] - p[x].a[i]) * (p[z].a[i] - p[x].a[i]);
  }
  return res;
}
int main() {
  int n = read();
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", &p[i].a[j]);
  for (int i = 1; i <= n; i++) {
    bool f = 0;
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      for (int k = j + 1; k <= n; k++) {
        if (k == i) continue;
        if (calc(i, j, k) > 0) {
          f = 1;
          break;
        }
      }
      if (f) break;
    }
    if (!f) ans.push_back(i);
  }
  cout << ans.size() << endl;
  for (int i = 0; i < (int)ans.size(); i++) cout << ans[i] << " ";
  if (ans.size()) puts("");
  return 0;
}

```