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.
#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
const double PI = acos(-1.0);
const double EPS = 1e-12;
struct Point {
  double a, b, c, d, e;
  double operator*(const Point &rhs) const {
    return a * rhs.a + b * rhs.b + c * rhs.c + d * rhs.d + e * rhs.e;
  }
} p[N];
bool ans[N];
Point Dir(int x, int y) {
  return Point{p[y].a - p[x].a, p[y].b - p[x].b, p[y].c - p[x].c,
               p[y].d - p[x].d, p[y].e - p[x].e};
}
double Cal(int x, int y) {
  return sqrt((p[x].a - p[y].a) * (p[x].a - p[y].a) +
              (p[x].b - p[y].b) * (p[x].b - p[y].b) +
              (p[x].c - p[y].c) * (p[x].c - p[y].c) +
              (p[x].d - p[y].d) * (p[x].d - p[y].d) +
              (p[x].e - p[y].e) * (p[x].e - p[y].e));
}
double Sol(int x, int y, int z) {
  double xy = Cal(x, y), xz = Cal(x, z);
  Point t1 = Dir(x, y), t2 = Dir(x, z);
  return t1 * t2 / (xy * xz);
}
void Check(int x, int y, int z) {
  double pxyxz, pyxyz, pzxzy;
  if (!ans[x]) {
    pxyxz = Sol(x, y, z);
    if (PI / 2 - acos(pxyxz) >= EPS) ans[x] = true;
  }
  if (!ans[y]) {
    pyxyz = Sol(y, x, z);
    if (PI / 2 - acos(pyxyz) >= EPS) ans[y] = true;
  }
  if (!ans[z]) {
    pzxzy = Sol(z, x, y);
    if (PI / 2 - acos(pzxzy) >= EPS) ans[z] = true;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%lf %lf %lf %lf %lf", &p[i].a, &p[i].b, &p[i].c, &p[i].d, &p[i].e);
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      for (int k = j + 1; k <= n; k++) Check(i, j, k);
  int cnt = 0;
  for (int i = 1; i <= n; i++) cnt += (ans[i] ? 0 : 1);
  printf("%d\n", cnt);
  for (int i = 1; i <= n; i++)
    if (!ans[i]) printf("%d\n", i);
  return 0;
}
