#include <bits/stdc++.h>
using namespace std;
struct Point {
  int x[5];
  void read() {
    for (int i = 0; i < 5; ++i) scanf("%d", &x[i]);
  }
} p[105];
long long operator*(const Point& a, const Point& b) {
  long long ret = 0;
  for (int i = 0; i < 5; ++i) ret += (long long)a.x[i] * b.x[i];
  return ret;
}
Point operator-(const Point& a, const Point& b) {
  Point ret;
  for (int i = 0; i < 5; ++i) ret.x[i] = a.x[i] - b.x[i];
  return ret;
}
int n;
bool fg;
vector<int> G;
bool pd(int x, int y, int z) { return (p[y] - p[x]) * (p[z] - p[x]) > 0; }
int main() {
  scanf("%d", &n);
  if (n > 100) {
    printf("0\n");
    return 0;
  }
  for (int i = 1; i <= n; ++i) p[i].read();
  for (int i = 1; i <= n; ++i) {
    fg = true;
    for (int j = 1; j <= n; ++j) {
      for (int k = j + 1; k <= n; ++k)
        if (i != j && i != k) {
          if (pd(i, j, k)) {
            fg = false;
            break;
          }
        }
      if (!fg) break;
    }
    if (fg) G.push_back(i);
  }
  printf("%d\n", G.size());
  for (int i = 0; i < G.size(); ++i) printf("%d\n", G[i]);
  return 0;
}

### Prompt

Your task is to create 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;
struct Point {
  int x[5];
  void read() {
    for (int i = 0; i < 5; ++i) scanf("%d", &x[i]);
  }
} p[105];
long long operator*(const Point& a, const Point& b) {
  long long ret = 0;
  for (int i = 0; i < 5; ++i) ret += (long long)a.x[i] * b.x[i];
  return ret;
}
Point operator-(const Point& a, const Point& b) {
  Point ret;
  for (int i = 0; i < 5; ++i) ret.x[i] = a.x[i] - b.x[i];
  return ret;
}
int n;
bool fg;
vector<int> G;
bool pd(int x, int y, int z) { return (p[y] - p[x]) * (p[z] - p[x]) > 0; }
int main() {
  scanf("%d", &n);
  if (n > 100) {
    printf("0\n");
    return 0;
  }
  for (int i = 1; i <= n; ++i) p[i].read();
  for (int i = 1; i <= n; ++i) {
    fg = true;
    for (int j = 1; j <= n; ++j) {
      for (int k = j + 1; k <= n; ++k)
        if (i != j && i != k) {
          if (pd(i, j, k)) {
            fg = false;
            break;
          }
        }
      if (!fg) break;
    }
    if (fg) G.push_back(i);
  }
  printf("%d\n", G.size());
  for (int i = 0; i < G.size(); ++i) printf("%d\n", G[i]);
  return 0;
}

```