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 double PI = acos(-1);
int n;
struct Node {
  int x[8];
} node[1005], t;
int cnt;
vector<int> ans;
vector<pair<double, int> > temp;
bool flag;
double dis(int id) {
  double x = t.x[1] * node[id].x[1] + t.x[2] * node[id].x[2] +
             t.x[3] * node[id].x[3] + t.x[4] * node[id].x[4] +
             t.x[5] * node[id].x[5];
  double a = t.x[1] * t.x[1] + t.x[2] * t.x[2] + t.x[3] * t.x[3] +
             t.x[4] * t.x[4] + t.x[5] * t.x[5];
  double b = node[id].x[1] * node[id].x[1] + node[id].x[2] * node[id].x[2] +
             node[id].x[3] * node[id].x[3] + node[id].x[4] * node[id].x[4] +
             node[id].x[5] * node[id].x[5];
  return acos(x / sqrt(a * b));
}
bool check(int p, int q, int r) {
  for (int i = 1; i <= 5; i++) {
    node[q].x[i] -= node[p].x[i];
    node[r].x[i] -= node[p].x[i];
  }
  int a = 0, b = 0, c = 0;
  for (int i = 1; i <= 5; i++) {
    a += node[q].x[i] * node[r].x[i];
  }
  for (int i = 1; i <= 5; i++) {
    node[q].x[i] += node[p].x[i];
    node[r].x[i] += node[p].x[i];
  }
  if (a > 0)
    return true;
  else
    return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d%d%d%d%d", &node[i].x[1], &node[i].x[2], &node[i].x[3],
          &node[i].x[4], &node[i].x[5]);
  t.x[1] = t.x[2] = t.x[3] = t.x[4] = t.x[5] = 2000;
  cnt = 0;
  for (int i = 1; i <= n; i++) {
    temp.clear();
    flag = true;
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      temp.push_back(pair<double, int>(dis(j), j));
    }
    sort(temp.begin(), temp.end());
    for (int j = 0; j < n - 1; j++) {
      int k = j + 1;
      if (k >= n - 1) k -= n - 1;
      if (j == k) break;
      if (check(i, temp[j].second, temp[k].second)) flag = false;
    }
    if (flag) {
      cnt++;
      ans.push_back(i);
    }
  }
  printf("%d\n", cnt);
  for (auto it : ans) printf("%d\n", it);
  return 0;
}
