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())
points = []
for p in range(n):
	points.append(list(map(int, input().split())))

def dot(x, y):
	res = 0
	for a, b in zip(x, y):
		res += a*b
	return res

def minus(x, y):
	res = []
	for a, b in zip(x, y):
		res.append(a-b)
	return res

indices = set(range(n))
if n <= 50:
	for x in range(n):
		for y in range(n):
			if x != y:
				for z in range(n):
					if z != y and z != x:

						if dot(minus(points[y], points[x]), minus(points[z], points[x])) > 0:
							indices.discard(x)
#							indices.discard(z)
	print(len(indices))
	for i in sorted(indices):
		print(i+1)						
else:
	print(0)
