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;
int good[1000], arr[5][1000];
const long double HALF_PI = 1.570796326;
double mult(double a[], double b[]) {
  double ans = 0;
  for (int i = 0; i < 5; i++) ans += a[i] * b[i];
  return ans;
}
bool isGood(int c, int l, int r) {
  double a[5], b[5];
  for (int i = 0; i < 5; i++) {
    a[i] = arr[i][l] - arr[i][c];
    b[i] = arr[i][r] - arr[i][c];
  }
  double alpha = acos(mult(a, b) / (sqrt(mult(b, b)) * sqrt(mult(a, a))));
  return alpha >= (HALF_PI);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    scanf("%d%d%d%d%d", &arr[0][i], &arr[1][i], &arr[2][i], &arr[3][i],
          &arr[4][i]);
  for (int i = 0; i < n; i++) {
    if (good[i]) continue;
    for (int j = i + 1; j < n; j++) {
      for (int k = j + 1; k < n; k++) {
        if (isGood(i, j, k)) {
          good[j] = 1;
          good[k] = 1;
        } else {
          if ((!good[j]) && (!isGood(j, i, k))) good[j] = 1;
          if ((!good[k]) && (!isGood(k, i, j))) good[j] = 1;
          good[i] = 1;
        }
      }
    }
  }
  for (int i = 0; i < n - 2; i++) {
    if (isGood(n - 1, n - 2, i)) {
      good[n - 2] = 1;
      good[i] = 1;
    } else
      good[n - 1] = 1;
    if (isGood(n - 2, n - 1, i)) {
      good[n - 1] = 1;
      good[i] = 1;
    } else
      good[n - 2] = 1;
  }
  int k = 0;
  for (int i = 0; i < n; i++) {
    if (!good[i]) k++;
  }
  printf("%d\n", k);
  for (int i = 0; i < n; i++) {
    if (!good[i]) printf("%d ", i + 1);
  }
  return 0;
}
