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 MXN = 1e3 + 10;
int n, m, r, t, res[MXN];
struct Point {
  int a, b, c, d, e;
};
Point A[MXN];
int Dist(Point X, Point Y) {
  int rr = 0;
  rr += X.a * Y.a;
  rr += X.b * Y.b;
  rr += X.c * Y.c;
  rr += X.d * Y.d;
  rr += X.e * Y.e;
  return (rr);
}
int Calc(int i, int j, int h) {
  Point X, Y;
  X.a = A[j].a - A[i].a;
  X.b = A[j].b - A[i].b;
  X.c = A[j].c - A[i].c;
  X.d = A[j].d - A[i].d;
  X.e = A[j].e - A[i].e;
  Y.a = A[h].a - A[i].a;
  Y.b = A[h].b - A[i].b;
  Y.c = A[h].c - A[i].c;
  Y.d = A[h].d - A[i].d;
  Y.e = A[h].e - A[i].e;
  long double rr = Dist(X, Y);
  long double xx = sqrt(Dist(X, X));
  long double yy = sqrt(Dist(Y, Y));
  if (rr / (xx * yy) > 0) return (0);
  return (1);
}
int main() {
  scanf("%d", &n);
  if (n >= 20) return printf("0"), 0;
  for (int i = 1; i <= n; i++)
    scanf("%d%d%d%d%d", &A[i].a, &A[i].b, &A[i].c, &A[i].d, &A[i].e),
        res[i] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      for (int h = j + 1; h <= n; h++) {
        if (i == j || i == h) continue;
        if (Calc(i, j, h) == 0) res[i] = 0;
      }
    }
  }
  for (int i = 1; i <= n; i++) r += (res[i]);
  printf("%d\n", r);
  for (int i = 1; i <= n; i++)
    if (res[i] == 1) printf("%d ", i);
  return (0);
}
