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 mod = 1e9 + 7;
const double eps = 1e-6;
const double pi = acos(-1);
const int inf = 0x7fffffff;
using namespace std;
struct node {
  int a[6];
  node operator-(const node& x) {
    node y;
    for (int i = 1; i <= 5; i++) y.a[i] = a[i] - x.a[i];
    return y;
  }
  double operator*(const node& x) {
    double res = 0;
    for (int i = 1; i <= 5; i++) res += 1.0 * a[i] * x.a[i];
    return res;
  }
} p[1005];
double mo(node& x) {
  double res = 0;
  for (int i = 1; i <= 5; i++) res += x.a[i] * x.a[i];
  return sqrt(res);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= 5; j++) scanf("%d", &p[i].a[j]);
  vector<int> G;
  for (int i = 1; i <= n; i++) {
    bool flag = true;
    for (int j = 1; j < n; j++) {
      if (j == i) continue;
      for (int k = j + 1; k <= n; k++) {
        if (k == i) continue;
        node ab = p[j] - p[i];
        node ac = p[k] - p[i];
        double res = acos((ab * ac) / (mo(ab) * mo(ac)));
        if (fabs(res) < pi / 2.0) {
          flag = false;
          goto L;
        }
      }
    }
    if (flag) G.push_back(i);
  L:;
  }
  printf("%d\n", G.size());
  for (int i = 0; i < G.size(); i++) printf("%d\n", G[i]);
  return 0;
}
