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 maxN = 1e3;
const long long inf = 1e18;
struct P {
  long long x[5];
  P operator-(P &rhs) {
    P res;
    for (int i = 0; i < 5; i++) res.x[i] = x[i] - rhs.x[i];
    return res;
  }
} pnt[maxN];
long long sq(long long x) { return x * x; }
long long dist(P &aa, P &bb) {
  long long ans = 0;
  for (int i = 0; i < 5; i++) ans += sq(aa.x[i] - bb.x[i]);
  return ans;
}
long long prod(P &aa, P &bb) {
  long long ans = 0;
  for (int i = 0; i < 5; i++) ans += aa.x[i] * bb.x[i];
  return ans;
}
int main() {
  int i, j, n;
  scanf("%d", &n);
  for (i = 0; i < n; i++)
    for (j = 0; j < 5; j++) scanf("%lld", &pnt[i].x[j]);
  if (n == 1) {
    puts("1");
    puts("1");
    return 0;
  }
  int a, b;
  long long md = inf;
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      long long d = dist(pnt[i], pnt[j]);
      if (d < md) md = d, a = i, b = j;
    }
  }
  vector<int> ans;
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      P u = pnt[i] - pnt[a];
      P v = pnt[j] - pnt[a];
      if (prod(u, v) > 0) goto ee;
    }
  }
  ans.push_back(a);
ee:
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      P u = pnt[i] - pnt[b];
      P v = pnt[j] - pnt[b];
      if (prod(u, v) > 0) goto end;
    }
  }
  ans.push_back(b);
end:
  printf("%d\n", ans.size());
  for (auto &elm : ans) printf("%d\n", elm + 1);
}
