#include <bits/stdc++.h>
using namespace std;
const int maxn = 55;
int n, x[maxn][5], vis[maxn];
inline void check(int a, int b, int c) {
  int reta = 0, retb = 0, retc = 0;
  for (int i = 0; i < 5; ++i) {
    reta += (x[b][i] - x[a][i]) * (x[c][i] - x[a][i]);
    retb += (x[a][i] - x[b][i]) * (x[c][i] - x[b][i]);
    retc += (x[a][i] - x[c][i]) * (x[b][i] - x[c][i]);
  }
  vis[a] |= (reta > 0);
  vis[b] |= (retb > 0);
  vis[c] |= (retc > 0);
}
int main() {
  scanf("%d", &n);
  if (n > 50) {
    puts("0");
    return 0;
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 5; ++j) scanf("%d", &x[i][j]);
  for (int i = 0; i < n; ++i)
    for (int j = i + 1; j < n; ++j)
      for (int k = j + 1; k < n; ++k) check(i, j, k);
  int res = 0;
  for (int i = 0; i < n; ++i) res += !vis[i];
  printf("%d\n", res);
  for (int i = 0; i < n; ++i)
    if (!vis[i]) printf("%d\n", i + 1);
  return 0;
}

### Prompt

Your challenge is to write 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 int maxn = 55;
int n, x[maxn][5], vis[maxn];
inline void check(int a, int b, int c) {
  int reta = 0, retb = 0, retc = 0;
  for (int i = 0; i < 5; ++i) {
    reta += (x[b][i] - x[a][i]) * (x[c][i] - x[a][i]);
    retb += (x[a][i] - x[b][i]) * (x[c][i] - x[b][i]);
    retc += (x[a][i] - x[c][i]) * (x[b][i] - x[c][i]);
  }
  vis[a] |= (reta > 0);
  vis[b] |= (retb > 0);
  vis[c] |= (retc > 0);
}
int main() {
  scanf("%d", &n);
  if (n > 50) {
    puts("0");
    return 0;
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 5; ++j) scanf("%d", &x[i][j]);
  for (int i = 0; i < n; ++i)
    for (int j = i + 1; j < n; ++j)
      for (int k = j + 1; k < n; ++k) check(i, j, k);
  int res = 0;
  for (int i = 0; i < n; ++i) res += !vis[i];
  printf("%d\n", res);
  for (int i = 0; i < n; ++i)
    if (!vis[i]) printf("%d\n", i + 1);
  return 0;
}

```