#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1010;
int N;
int X[MAXN][5];
int dot(int a, int b, int c) {
  int d = 0;
  for (int i = 0; i < 5; i++) {
    d += (X[b][i] - X[a][i]) * (X[c][i] - X[a][i]);
  }
  return d;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", &X[i][j]);
    }
  }
  if (N >= 500) {
    puts("0");
    return 0;
  }
  vector<int> ans;
  for (int i = 0; i < N; i++) {
    bool good = true;
    for (int j = 0; j < N; j++) {
      if (j != i) {
        for (int k = j + 1; k < N; k++) {
          if (k != i && k != j) {
            int d = dot(i, j, k);
            if (d > 0) {
              good = false;
              break;
            }
          }
        }
      }
      if (!good) {
        break;
      }
    }
    if (good) {
      ans.push_back(i);
    }
  }
  printf("%lu\n", ans.size());
  for (int x : ans) {
    printf("%d\n", x + 1);
  }
}

### Prompt

In CPP, your task is to solve the following problem:

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;
const int MAXN = 1010;
int N;
int X[MAXN][5];
int dot(int a, int b, int c) {
  int d = 0;
  for (int i = 0; i < 5; i++) {
    d += (X[b][i] - X[a][i]) * (X[c][i] - X[a][i]);
  }
  return d;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", &X[i][j]);
    }
  }
  if (N >= 500) {
    puts("0");
    return 0;
  }
  vector<int> ans;
  for (int i = 0; i < N; i++) {
    bool good = true;
    for (int j = 0; j < N; j++) {
      if (j != i) {
        for (int k = j + 1; k < N; k++) {
          if (k != i && k != j) {
            int d = dot(i, j, k);
            if (d > 0) {
              good = false;
              break;
            }
          }
        }
      }
      if (!good) {
        break;
      }
    }
    if (good) {
      ans.push_back(i);
    }
  }
  printf("%lu\n", ans.size());
  for (int x : ans) {
    printf("%d\n", x + 1);
  }
}

```