#include <bits/stdc++.h>
using namespace std;
int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
int n, m, i, j, k, s, o;
int a[1005][7];
int pr[1005], cnt;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i][1] = read();
    a[i][2] = read();
    a[i][3] = read();
    a[i][4] = read();
    a[i][5] = read();
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j < n; j++) {
      if (i != j) {
        for (int k = j + 1; k <= n; k++) {
          if (k != i) {
            long long ansi = 0;
            for (int l = 1; l <= 5; l++) {
              ansi += (a[j][l] - a[i][l]) * (a[k][l] - a[i][l]);
            }
            if (ansi > 0) goto BROKEN;
          }
        }
      }
    }
    pr[++cnt] = i;
  BROKEN:;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d\n", pr[i]);
  }
  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 read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
int n, m, i, j, k, s, o;
int a[1005][7];
int pr[1005], cnt;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i][1] = read();
    a[i][2] = read();
    a[i][3] = read();
    a[i][4] = read();
    a[i][5] = read();
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j < n; j++) {
      if (i != j) {
        for (int k = j + 1; k <= n; k++) {
          if (k != i) {
            long long ansi = 0;
            for (int l = 1; l <= 5; l++) {
              ansi += (a[j][l] - a[i][l]) * (a[k][l] - a[i][l]);
            }
            if (ansi > 0) goto BROKEN;
          }
        }
      }
    }
    pr[++cnt] = i;
  BROKEN:;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d\n", pr[i]);
  }
  return 0;
}

```