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;
struct node {
  int a[6];
} N[1000 + 5];
vector<int> V;
int dis[1000 + 5][1000 + 5][10];
bool judge(int x, int y, int z) {
  double xy = 0;
  double _x = 0;
  double _y = 0;
  for (int i = 1; i <= 5; i++) {
    xy += dis[y][x][i] * dis[z][x][i];
    _x += pow(dis[y][x][i], 2);
    _y += pow(dis[z][x][i], 2);
  }
  _x = sqrt(_x);
  _y = sqrt(_y);
  double final = xy / (_x * _y);
  if (final > 0)
    return true;
  else
    return false;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 5; j++) cin >> N[i].a[j];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      for (int k = 1; k <= 5; k++) dis[i][j][k] = N[i].a[k] - N[j].a[k];
    }
  }
  int sum = 0;
  V.clear();
  for (int i = 1; i <= n; i++) {
    int flag = 1;
    for (int j = 1; j <= n; j++) {
      if (j == i) continue;
      for (int k = j + 1; k <= n; k++) {
        if (judge(i, j, k)) {
          flag = 0;
        }
        if (flag == 0) break;
      }
      if (flag == 0) break;
    }
    if (flag) {
      sum++;
      V.push_back(i);
    }
  }
  cout << sum << endl;
  for (int i = 0; i < V.size(); i++) {
    cout << V[i] << endl;
  }
}
