#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
struct node {
  int a[7];
} p[N];
int n, ans[N], top = 0;
node operator-(node p1, node p2) {
  node p3;
  for (int i = 1; i <= 5; i++) p3.a[i] = p1.a[i] - p2.a[i];
  return p3;
}
int operator*(node p1, node p2) {
  int i, j = 0;
  for (i = 1; i <= 5; i++) j += p1.a[i] * p2.a[i];
  return j;
}
void Solve() {
  int i, j, k, tag;
  for (i = 1; i <= n; i++) {
    tag = 1;
    for (j = 1; j <= n && tag; j++) {
      if (j == i) continue;
      for (k = 1; k <= n && tag; k++) {
        if (k == i || k == j) continue;
        if ((p[j] - p[i]) * (p[k] - p[i]) > 0) {
          tag = 0;
        }
      }
    }
    if (tag) ans[++top] = i;
  }
  printf("%d", top);
  if (top) {
    printf("\n");
    for (i = 1; i <= top; i++) printf("%d ", ans[i]);
  }
}
int main() {
  scanf("%d", &n);
  int i, j;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= 5; j++) scanf("%d", &p[i].a[j]);
  if (n <= 400)
    Solve();
  else
    printf("0");
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 N = 1005;
struct node {
  int a[7];
} p[N];
int n, ans[N], top = 0;
node operator-(node p1, node p2) {
  node p3;
  for (int i = 1; i <= 5; i++) p3.a[i] = p1.a[i] - p2.a[i];
  return p3;
}
int operator*(node p1, node p2) {
  int i, j = 0;
  for (i = 1; i <= 5; i++) j += p1.a[i] * p2.a[i];
  return j;
}
void Solve() {
  int i, j, k, tag;
  for (i = 1; i <= n; i++) {
    tag = 1;
    for (j = 1; j <= n && tag; j++) {
      if (j == i) continue;
      for (k = 1; k <= n && tag; k++) {
        if (k == i || k == j) continue;
        if ((p[j] - p[i]) * (p[k] - p[i]) > 0) {
          tag = 0;
        }
      }
    }
    if (tag) ans[++top] = i;
  }
  printf("%d", top);
  if (top) {
    printf("\n");
    for (i = 1; i <= top; i++) printf("%d ", ans[i]);
  }
}
int main() {
  scanf("%d", &n);
  int i, j;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= 5; j++) scanf("%d", &p[i].a[j]);
  if (n <= 400)
    Solve();
  else
    printf("0");
  return 0;
}

```