#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
inline void read(long long &x) {
  x = 0;
  long long f = 1;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s ^ 48);
    s = getchar();
  }
  x *= f;
}
inline long long Max(long long x, long long y) { return x > y ? x : y; }
inline long long Min(long long x, long long y) { return x < y ? x : y; }
inline long long Abs(long long x) {
  if (x > 0) return x;
  return -x;
}
long long a[400005][6], Good[400005];
int main() {
  long long n;
  read(n);
  for (register int i = 1; i <= n; ++i)
    for (register int j = 1; j <= 5; ++j) read(a[i][j]);
  if (n > 11) return !printf("0");
  long long tot = 0;
  for (register int i = 1; i <= n; ++i) {
    for (register int j = 1; j <= n; ++j) {
      if (i != j) {
        for (register int k = 1; k <= n; ++k)
          if (i != k && j != k) {
            long long JD = 0;
            for (register int l = 1; l <= 5; ++l)
              JD += (a[j][l] - a[i][l]) * (a[k][l] - a[i][l]);
            if (JD > 0) goto end;
          }
      }
    }
    Good[++tot] = i;
  end:;
  }
  printf("%lld\n", tot);
  for (register int i = 1; i <= tot; ++i) printf("%lld\n", Good[i]);
}

### Prompt

Create a solution in Cpp for 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>
#pragma GCC optimize(2)
using namespace std;
inline void read(long long &x) {
  x = 0;
  long long f = 1;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s ^ 48);
    s = getchar();
  }
  x *= f;
}
inline long long Max(long long x, long long y) { return x > y ? x : y; }
inline long long Min(long long x, long long y) { return x < y ? x : y; }
inline long long Abs(long long x) {
  if (x > 0) return x;
  return -x;
}
long long a[400005][6], Good[400005];
int main() {
  long long n;
  read(n);
  for (register int i = 1; i <= n; ++i)
    for (register int j = 1; j <= 5; ++j) read(a[i][j]);
  if (n > 11) return !printf("0");
  long long tot = 0;
  for (register int i = 1; i <= n; ++i) {
    for (register int j = 1; j <= n; ++j) {
      if (i != j) {
        for (register int k = 1; k <= n; ++k)
          if (i != k && j != k) {
            long long JD = 0;
            for (register int l = 1; l <= 5; ++l)
              JD += (a[j][l] - a[i][l]) * (a[k][l] - a[i][l]);
            if (JD > 0) goto end;
          }
      }
    }
    Good[++tot] = i;
  end:;
  }
  printf("%lld\n", tot);
  for (register int i = 1; i <= tot; ++i) printf("%lld\n", Good[i]);
}

```