#include <bits/stdc++.h>
using namespace std;
const int MAXB = 1e5;
char buf[MAXB], *cp = buf;
inline void rd(int &x) {
  x = 0;
  int f = 1;
  while (*cp < '0' || '9' < *cp) {
    if (*cp == '-') f = -1;
    cp++;
  }
  while ('0' <= *cp && *cp <= '9') x = (x << 3) + (x << 1) + *cp - 48, cp++;
  x *= f;
}
const int MAXN = 1010;
int n, cnt;
int t[MAXN][5], ans[MAXN];
int main() {
  fread(buf, 1, MAXB, stdin);
  rd(n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) rd(t[i][j]);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++)
      if (j != i)
        for (int k = j + 1; k < n; k++)
          if (k != i) {
            int tmp = 0;
            for (int l = 0; l < 5; l++)
              tmp += (t[j][l] - t[i][l]) * (t[k][l] - t[i][l]);
            if (tmp > 0) goto nxt;
          }
    ans[cnt++] = i;
  nxt:;
  }
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) printf("%d ", ans[i] + 1);
  printf("\n");
  return 0;
}

### 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;
const int MAXB = 1e5;
char buf[MAXB], *cp = buf;
inline void rd(int &x) {
  x = 0;
  int f = 1;
  while (*cp < '0' || '9' < *cp) {
    if (*cp == '-') f = -1;
    cp++;
  }
  while ('0' <= *cp && *cp <= '9') x = (x << 3) + (x << 1) + *cp - 48, cp++;
  x *= f;
}
const int MAXN = 1010;
int n, cnt;
int t[MAXN][5], ans[MAXN];
int main() {
  fread(buf, 1, MAXB, stdin);
  rd(n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) rd(t[i][j]);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++)
      if (j != i)
        for (int k = j + 1; k < n; k++)
          if (k != i) {
            int tmp = 0;
            for (int l = 0; l < 5; l++)
              tmp += (t[j][l] - t[i][l]) * (t[k][l] - t[i][l]);
            if (tmp > 0) goto nxt;
          }
    ans[cnt++] = i;
  nxt:;
  }
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) printf("%d ", ans[i] + 1);
  printf("\n");
  return 0;
}

```