#include <bits/stdc++.h>
long long expow(long long a, long long b, long long p) {
  long long v = 1;
  for (; b; b >>= 1, a = a * a % p)
    if (b & 1) v = v * a % p;
  return v;
}
long long inv(long long a, long long p) { return expow(a, p - 2, p); }
using namespace std;
const int N = 1000 + 11;
const int M = 1000000007;
struct node {
  int a, b, c, d, e;
} p[N];
node sub(node a, node b) {
  return (node){a.a - b.a, a.b - b.b, a.c - b.c, a.d - b.d, a.e - b.e};
}
double mix(node a, node b) {
  return 1.0 * a.a * b.a + a.b * b.b + a.c * b.c + a.d * b.d + a.e * b.e;
}
vector<int> ans;
int main() {
  int k;
  scanf("%d", &k);
  for (int i = 0; i < k; i++) {
    scanf("%d%d%d%d%d", &p[i].a, &p[i].b, &p[i].c, &p[i].d, &p[i].e);
  }
  for (int j = 0; j < k; j++) {
    bool f = false;
    for (int i = 0; i < k; i++)
      if (i != j) {
        for (int l = 0; l < k; l++)
          if (l != i && l != j) {
            node cha = sub(p[j], p[i]), cha2 = sub(p[j], p[l]);
            double ding = mix(cha, cha2),
                   di = sqrt(mix(cha, cha)) * sqrt(mix(cha2, cha2));
            double xx = ding / di;
            if (xx > 0 && xx <= 1) {
              f = true;
              break;
            }
          }
        if (f) break;
      }
    if (!f) ans.push_back(j + 1);
  }
  int sz = ans.size();
  printf("%d\n", sz);
  for (int j = 0; j < sz; j++) printf("%d\n", ans[j]);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
long long expow(long long a, long long b, long long p) {
  long long v = 1;
  for (; b; b >>= 1, a = a * a % p)
    if (b & 1) v = v * a % p;
  return v;
}
long long inv(long long a, long long p) { return expow(a, p - 2, p); }
using namespace std;
const int N = 1000 + 11;
const int M = 1000000007;
struct node {
  int a, b, c, d, e;
} p[N];
node sub(node a, node b) {
  return (node){a.a - b.a, a.b - b.b, a.c - b.c, a.d - b.d, a.e - b.e};
}
double mix(node a, node b) {
  return 1.0 * a.a * b.a + a.b * b.b + a.c * b.c + a.d * b.d + a.e * b.e;
}
vector<int> ans;
int main() {
  int k;
  scanf("%d", &k);
  for (int i = 0; i < k; i++) {
    scanf("%d%d%d%d%d", &p[i].a, &p[i].b, &p[i].c, &p[i].d, &p[i].e);
  }
  for (int j = 0; j < k; j++) {
    bool f = false;
    for (int i = 0; i < k; i++)
      if (i != j) {
        for (int l = 0; l < k; l++)
          if (l != i && l != j) {
            node cha = sub(p[j], p[i]), cha2 = sub(p[j], p[l]);
            double ding = mix(cha, cha2),
                   di = sqrt(mix(cha, cha)) * sqrt(mix(cha2, cha2));
            double xx = ding / di;
            if (xx > 0 && xx <= 1) {
              f = true;
              break;
            }
          }
        if (f) break;
      }
    if (!f) ans.push_back(j + 1);
  }
  int sz = ans.size();
  printf("%d\n", sz);
  for (int j = 0; j < sz; j++) printf("%d\n", ans[j]);
  return 0;
}

```