#include <bits/stdc++.h>
using namespace std;
int a[1010][5];
bool isac(int l[5], int r[5]) {
  int p = 0;
  for (int k = 0; k < 5; k++) {
    p += l[k] * r[k];
  }
  return p > 0;
}
int main() {
  int n;
  scanf("%d", &(n));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", &(a[i][j]));
    }
  }
  if (n > 50) {
    printf("0\n");
    return 0;
  }
  vector<int> v;
  for (int i = 0; i < n; i++) {
    int ok = 1, good = 1;
    for (int j = 0; j < n && ok; j++) {
      if (j == i) continue;
      for (int k = 0; k < n && ok; k++) {
        if (k == i || k == j) continue;
        int l[5], r[5];
        for (int o = 0; o < 5; o++) {
          l[o] = a[j][o] - a[i][o];
          r[o] = a[k][o] - a[i][o];
        }
        if (isac(l, r)) {
          good = 0;
          ok = 0;
        }
      }
    }
    if (good) v.push_back(i);
  }
  printf("%d\n", (int)v.size());
  sort(v.begin(), v.end());
  for (int i : v) {
    printf("%d\n", i + 1);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int a[1010][5];
bool isac(int l[5], int r[5]) {
  int p = 0;
  for (int k = 0; k < 5; k++) {
    p += l[k] * r[k];
  }
  return p > 0;
}
int main() {
  int n;
  scanf("%d", &(n));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", &(a[i][j]));
    }
  }
  if (n > 50) {
    printf("0\n");
    return 0;
  }
  vector<int> v;
  for (int i = 0; i < n; i++) {
    int ok = 1, good = 1;
    for (int j = 0; j < n && ok; j++) {
      if (j == i) continue;
      for (int k = 0; k < n && ok; k++) {
        if (k == i || k == j) continue;
        int l[5], r[5];
        for (int o = 0; o < 5; o++) {
          l[o] = a[j][o] - a[i][o];
          r[o] = a[k][o] - a[i][o];
        }
        if (isac(l, r)) {
          good = 0;
          ok = 0;
        }
      }
    }
    if (good) v.push_back(i);
  }
  printf("%d\n", (int)v.size());
  sort(v.begin(), v.end());
  for (int i : v) {
    printf("%d\n", i + 1);
  }
  return 0;
}

```