#include <bits/stdc++.h>
using namespace std;
double eps = 0.0000001;
struct vec {
  double tg[5];
  bool operator<(const vec &a) const {
    for (int i = 0; i < 5; i++)
      if (abs(tg[i] - a.tg[i]) > eps) return tg[i] < a.tg[i];
    return false;
  }
} p[1005];
bool judge(vec a, vec b) {
  double ans = 0;
  for (int i = 0; i < 5; i++) ans += a.tg[i] * b.tg[i];
  if (ans > eps) return 1;
  return 0;
}
int x[5][1005];
int n;
bool fl[1005];
int st[1005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", &x[j][i]);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    int ncnt = 0;
    for (int j = 1; j <= n; j++)
      if (i == j)
        continue;
      else {
        for (int k = 0; k < 5; k++) p[ncnt].tg[k] = x[k][j] - x[k][i];
        ncnt++;
      }
    sort(p, p + ncnt);
    int cts = 0;
    for (int j = 0; j < ncnt; j++) st[cts++] = j;
    if (cts > 100)
      fl[i] = 1;
    else {
      for (int j = 0; j < cts; j++)
        for (int k = j + 1; k < cts; k++)
          if (judge(p[j], p[k])) fl[i] = 1;
    }
    if (!fl[i]) cnt++;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= n; i++)
    if (!fl[i]) printf("%d\n", 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>
using namespace std;
double eps = 0.0000001;
struct vec {
  double tg[5];
  bool operator<(const vec &a) const {
    for (int i = 0; i < 5; i++)
      if (abs(tg[i] - a.tg[i]) > eps) return tg[i] < a.tg[i];
    return false;
  }
} p[1005];
bool judge(vec a, vec b) {
  double ans = 0;
  for (int i = 0; i < 5; i++) ans += a.tg[i] * b.tg[i];
  if (ans > eps) return 1;
  return 0;
}
int x[5][1005];
int n;
bool fl[1005];
int st[1005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", &x[j][i]);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    int ncnt = 0;
    for (int j = 1; j <= n; j++)
      if (i == j)
        continue;
      else {
        for (int k = 0; k < 5; k++) p[ncnt].tg[k] = x[k][j] - x[k][i];
        ncnt++;
      }
    sort(p, p + ncnt);
    int cts = 0;
    for (int j = 0; j < ncnt; j++) st[cts++] = j;
    if (cts > 100)
      fl[i] = 1;
    else {
      for (int j = 0; j < cts; j++)
        for (int k = j + 1; k < cts; k++)
          if (judge(p[j], p[k])) fl[i] = 1;
    }
    if (!fl[i]) cnt++;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= n; i++)
    if (!fl[i]) printf("%d\n", i);
  return 0;
}

```