#include <bits/stdc++.h>
using namespace std;
int vis[1005];
struct node {
  int a, b, c, e, d;
  friend node operator-(node x, node y) {
    node z;
    z.a = x.a - y.a;
    z.b = x.b - y.b;
    z.c = x.c - y.c;
    z.d = x.d - y.d;
    z.e = x.e - y.e;
    return z;
  }
};
node data[1005];
int angle(node x, node y) {
  return (x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e);
}
int main() {
  int n;
  scanf("%d", &n);
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < n; ++i)
    scanf("%d%d%d%d%d", &data[i].a, &data[i].b, &data[i].c, &data[i].d,
          &data[i].e);
  if (n > 300) {
    printf("0\n");
    return 0;
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      for (int k = 0; k < n; ++k) {
        if (i != j && i != k && j != k)
          if (angle(data[j] - data[i], data[k] - data[i]) > 0) {
            vis[i] = 1;
          }
      }
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    if (vis[i] == 0) cnt++;
  }
  printf("%d\n", cnt);
  for (int i = 0; i < n; ++i) {
    if (vis[i] == 0) printf("%d\n", i + 1);
  }
  return 0;
}

### Prompt

Please formulate 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;
int vis[1005];
struct node {
  int a, b, c, e, d;
  friend node operator-(node x, node y) {
    node z;
    z.a = x.a - y.a;
    z.b = x.b - y.b;
    z.c = x.c - y.c;
    z.d = x.d - y.d;
    z.e = x.e - y.e;
    return z;
  }
};
node data[1005];
int angle(node x, node y) {
  return (x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e);
}
int main() {
  int n;
  scanf("%d", &n);
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < n; ++i)
    scanf("%d%d%d%d%d", &data[i].a, &data[i].b, &data[i].c, &data[i].d,
          &data[i].e);
  if (n > 300) {
    printf("0\n");
    return 0;
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      for (int k = 0; k < n; ++k) {
        if (i != j && i != k && j != k)
          if (angle(data[j] - data[i], data[k] - data[i]) > 0) {
            vis[i] = 1;
          }
      }
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    if (vis[i] == 0) cnt++;
  }
  printf("%d\n", cnt);
  for (int i = 0; i < n; ++i) {
    if (vis[i] == 0) printf("%d\n", i + 1);
  }
  return 0;
}

```