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 Point {
  double a[5];
  Point() {
    for (int i = 0; i < 5; i++) a[0] = 0;
  }
};
double mul(Point &a, Point &b) {
  double ret = 0;
  for (int i = 0; i < 5; i++) {
    ret += a.a[i] * b.a[i];
  }
  return ret;
}
double getC(Point &a, Point &b) {
  double t1 = mul(a, b);
  return t1;
}
const int maxn = 1000 + 10;
Point p[maxn];
vector<int> res;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%lf", &p[i].a[j]);
    }
  }
  double PI = acos(-1) / 2;
  for (int i = 0; i < n; i++) {
    bool flag = 1;
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      for (int k = j + 1; k < n; k++) {
        Point v1, v2;
        for (int w = 0; w < 5; w++) {
          v1.a[w] = p[j].a[w] - p[i].a[w];
          v2.a[w] = p[k].a[w] - p[i].a[w];
        }
        if (getC(v1, v2) > 0) {
          flag = 0;
          break;
        }
      }
      if (flag == 0) break;
    }
    if (flag) res.push_back(i);
  }
  printf("%d\n", (int)res.size());
  for (int i = 0; i < res.size(); i++) {
    printf("%d\n", res[i] + 1);
  }
  return 0;
}
