#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read1(T &m) {
  T x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  m = x * f;
}
template <typename T>
void read2(T &a, T &b) {
  read1(a);
  read1(b);
}
template <typename T>
void read3(T &a, T &b, T &c) {
  read1(a);
  read1(b);
  read1(c);
}
template <typename T>
void out(T a) {
  if (a > 9) out(a / 10);
  putchar(a % 10 + '0');
}
template <typename T>
void outn(T a) {
  if (a > 9) out(a / 10);
  putchar(a % 10 + '0');
  puts("");
}
struct P {
  int a, b, c, d, e;
} poi[1005];
bool dianji(P a, P b, P c) {
  int i = (b.a - a.a) * (c.a - a.a) + (b.b - a.b) * (c.b - a.b) +
          (b.c - a.c) * (c.c - a.c) + (b.d - a.d) * (c.d - a.d) +
          (b.e - a.e) * (c.e - a.e);
  if (i > 0) return false;
  return true;
}
int ret[1005];
int main() {
  int len;
  while (scanf("%d", &(len)) != EOF) {
    for (int i = (0); i < (len); i++)
      scanf("%d%d%d%d%d", &poi[i].a, &poi[i].b, &poi[i].c, &poi[i].d,
            &poi[i].e);
    int ans = 0;
    bool flag;
    for (int i = (0); i < (len); i++) {
      flag = true;
      for (int j = (0); j < (len); j++) {
        if (!flag) break;
        if (j == i) continue;
        for (int k = (0); k < (len); k++) {
          if (!flag) break;
          if (j == i || j == k) continue;
          if (dianji(poi[i], poi[j], poi[k]))
            continue;
          else {
            flag = false;
          }
        }
      }
      if (flag) ret[ans++] = i + 1;
    }
    printf("%d\n", ans);
    for (int i = (0); i < (ans); i++) printf("%d ", ret[i]);
  }
  return 0;
}

### 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>
using namespace std;
template <typename T>
void read1(T &m) {
  T x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  m = x * f;
}
template <typename T>
void read2(T &a, T &b) {
  read1(a);
  read1(b);
}
template <typename T>
void read3(T &a, T &b, T &c) {
  read1(a);
  read1(b);
  read1(c);
}
template <typename T>
void out(T a) {
  if (a > 9) out(a / 10);
  putchar(a % 10 + '0');
}
template <typename T>
void outn(T a) {
  if (a > 9) out(a / 10);
  putchar(a % 10 + '0');
  puts("");
}
struct P {
  int a, b, c, d, e;
} poi[1005];
bool dianji(P a, P b, P c) {
  int i = (b.a - a.a) * (c.a - a.a) + (b.b - a.b) * (c.b - a.b) +
          (b.c - a.c) * (c.c - a.c) + (b.d - a.d) * (c.d - a.d) +
          (b.e - a.e) * (c.e - a.e);
  if (i > 0) return false;
  return true;
}
int ret[1005];
int main() {
  int len;
  while (scanf("%d", &(len)) != EOF) {
    for (int i = (0); i < (len); i++)
      scanf("%d%d%d%d%d", &poi[i].a, &poi[i].b, &poi[i].c, &poi[i].d,
            &poi[i].e);
    int ans = 0;
    bool flag;
    for (int i = (0); i < (len); i++) {
      flag = true;
      for (int j = (0); j < (len); j++) {
        if (!flag) break;
        if (j == i) continue;
        for (int k = (0); k < (len); k++) {
          if (!flag) break;
          if (j == i || j == k) continue;
          if (dianji(poi[i], poi[j], poi[k]))
            continue;
          else {
            flag = false;
          }
        }
      }
      if (flag) ret[ans++] = i + 1;
    }
    printf("%d\n", ans);
    for (int i = (0); i < (ans); i++) printf("%d ", ret[i]);
  }
  return 0;
}

```