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>
const int MAXN = 1005;
using namespace std;
struct Point {
  int a, b, c, d, e;
  Point() {}
  Point(int a, int b, int c, int d, int e) : a(a), b(b), c(c), d(d), e(e) {}
} p[MAXN];
Point operator-(Point x, Point y) {
  return Point(x.a - y.a, x.b - y.b, x.c - y.c, x.d - y.d, x.e - y.e);
}
int dot(Point x, Point y) {
  return x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d%d%d%d%d", &p[i].a, &p[i].b, &p[i].c, &p[i].d, &p[i].e);
  if (n > 11) {
    printf("0\n");
    return 0;
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    bool flag = true;
    for (int j = 1; j <= n; j++)
      if (j != i)
        for (int k = j + 1; k <= n; k++)
          if (k != i)
            if (dot(p[j] - p[i], p[k] - p[i]) > 0) flag = false;
    if (flag) ans.push_back(i);
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  return 0;
}
