#include <bits/stdc++.h>
using namespace std;
const double P2 = acos(0);
struct Point {
  int a, b, c, d, e;
  Point(int a, int b, int c, int d, int e) : a(a), b(b), c(c), d(d), e(e) {}
};
Point vb(Point p1, Point p2) {
  return Point(p2.a - p1.a, p2.b - p1.b, p2.c - p1.c, p2.d - p1.d, p2.e - p1.e);
}
double dot(Point p1, Point p2) {
  return p1.a * p2.a + p1.b * p2.b + p1.c * p2.c + p1.d * p2.d + p1.e * p2.e;
}
double angle(Point p1, Point p2, Point p3) {
  Point v1 = vb(p1, p2);
  Point v2 = vb(p1, p3);
  return acos(dot(v1, v2) / sqrt(dot(v1, v1)) / sqrt(dot(v2, v2)));
}
int dsign(double d) {
  if (abs(d) < 1e-6) return 0;
  if (d > 0) return 1;
  return -1;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<Point> v;
  for (int i = 0; i < n; i++) {
    int a, b, c, d, e;
    scanf("%d%d%d%d%d", &a, &b, &c, &d, &e);
    v.push_back(Point(a, b, c, d, e));
  }
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    bool good = true;
    for (int j = 0; j < n; j++) {
      for (int k = j + 1; k < n; k++) {
        if (i != j && i != k) {
          double ang = angle(v[i], v[j], v[k]);
          if (dsign(ang - P2) < 0) {
            good = false;
            break;
          }
        }
      }
      if (!good) break;
    }
    if (good) ans.push_back(i + 1);
  }
  printf("%d\n", (int)ans.size());
  for (int i : ans) {
    printf("%d\n", i);
  }
  return 0;
}

### Prompt

Generate 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 double P2 = acos(0);
struct Point {
  int a, b, c, d, e;
  Point(int a, int b, int c, int d, int e) : a(a), b(b), c(c), d(d), e(e) {}
};
Point vb(Point p1, Point p2) {
  return Point(p2.a - p1.a, p2.b - p1.b, p2.c - p1.c, p2.d - p1.d, p2.e - p1.e);
}
double dot(Point p1, Point p2) {
  return p1.a * p2.a + p1.b * p2.b + p1.c * p2.c + p1.d * p2.d + p1.e * p2.e;
}
double angle(Point p1, Point p2, Point p3) {
  Point v1 = vb(p1, p2);
  Point v2 = vb(p1, p3);
  return acos(dot(v1, v2) / sqrt(dot(v1, v1)) / sqrt(dot(v2, v2)));
}
int dsign(double d) {
  if (abs(d) < 1e-6) return 0;
  if (d > 0) return 1;
  return -1;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<Point> v;
  for (int i = 0; i < n; i++) {
    int a, b, c, d, e;
    scanf("%d%d%d%d%d", &a, &b, &c, &d, &e);
    v.push_back(Point(a, b, c, d, e));
  }
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    bool good = true;
    for (int j = 0; j < n; j++) {
      for (int k = j + 1; k < n; k++) {
        if (i != j && i != k) {
          double ang = angle(v[i], v[j], v[k]);
          if (dsign(ang - P2) < 0) {
            good = false;
            break;
          }
        }
      }
      if (!good) break;
    }
    if (good) ans.push_back(i + 1);
  }
  printf("%d\n", (int)ans.size());
  for (int i : ans) {
    printf("%d\n", i);
  }
  return 0;
}

```