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.
n = int(input())
if n > 11:
  for _ in range(n):
    input()
  print(0)

else:
  L = [tuple(map(int,input().split())) for _ in range(n)]
  good = [True]*n

  from itertools import combinations,chain

  dot = lambda i,j: sum(L[i][k]*L[j][k] for k in range(5))

  memo = [[dot(i,j) for j in range(n)]for i in range(n)]

  for i in range(n):
    c2 = memo[i][i]
    for j,k in combinations(chain(range(i),range(i+1,n)), r=2):
      if memo[j][k] > memo[j][i] + memo[k][i] - c2:
        good[i] = False
        break

  print(sum(good))
  print(' '.join(map(str,(i+1 for i,v in enumerate(good) if v) )))