#include <bits/stdc++.h>
using namespace std;
int ans[1010];
int cnt;
int vis[1010];
struct P {
  int a[6];
  P() {}
  P(int a1, int a2, int a3, int a4, int a5) {
    a[1] = a1;
    a[2] = a2;
    a[3] = a3;
    a[4] = a4;
    a[5] = a5;
  }
  int& operator[](int i) { return a[i]; }
  void read() { scanf("%d%d%d%d%d", &a[1], &a[2], &a[3], &a[4], &a[5]); }
  P operator-(P p) {
    return P(a[1] - p[1], a[2] - p[2], a[3] - p[3], a[4] - p[4], a[5] - p[5]);
  }
};
P c[1010];
int calc(P p1, P p2) {
  int ret = 0;
  int k;
  for (k = 1; k <= 5; k++) ret += p1[k] * p2[k];
  return ret;
}
int main() {
  int n, i, j, k, flag;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) c[i].read();
  for (i = 1; i <= n; i++) {
    if (vis[i]) continue;
    flag = 1;
    for (j = 1; j <= n; j++) {
      for (k = j + 1; k <= n; k++) {
        if (j != i && k != i) {
          if (calc(c[j] - c[i], c[k] - c[i]) > 0) {
            flag = 0;
            break;
          } else {
            vis[j] = 1;
            vis[k] = 1;
          }
        }
        if (!flag) break;
      }
      if (!flag) break;
    }
    if (flag) ans[++cnt] = i;
  }
  printf("%d\n", cnt);
  for (i = 1; i <= cnt; i++) printf("%d\n", ans[i]);
}

### 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;
int ans[1010];
int cnt;
int vis[1010];
struct P {
  int a[6];
  P() {}
  P(int a1, int a2, int a3, int a4, int a5) {
    a[1] = a1;
    a[2] = a2;
    a[3] = a3;
    a[4] = a4;
    a[5] = a5;
  }
  int& operator[](int i) { return a[i]; }
  void read() { scanf("%d%d%d%d%d", &a[1], &a[2], &a[3], &a[4], &a[5]); }
  P operator-(P p) {
    return P(a[1] - p[1], a[2] - p[2], a[3] - p[3], a[4] - p[4], a[5] - p[5]);
  }
};
P c[1010];
int calc(P p1, P p2) {
  int ret = 0;
  int k;
  for (k = 1; k <= 5; k++) ret += p1[k] * p2[k];
  return ret;
}
int main() {
  int n, i, j, k, flag;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) c[i].read();
  for (i = 1; i <= n; i++) {
    if (vis[i]) continue;
    flag = 1;
    for (j = 1; j <= n; j++) {
      for (k = j + 1; k <= n; k++) {
        if (j != i && k != i) {
          if (calc(c[j] - c[i], c[k] - c[i]) > 0) {
            flag = 0;
            break;
          } else {
            vis[j] = 1;
            vis[k] = 1;
          }
        }
        if (!flag) break;
      }
      if (!flag) break;
    }
    if (flag) ans[++cnt] = i;
  }
  printf("%d\n", cnt);
  for (i = 1; i <= cnt; i++) printf("%d\n", ans[i]);
}

```