#include <bits/stdc++.h>
using namespace std;
int scalar(vector<int> &W1, vector<int> &W2) {
  int sc = 0;
  for (int i = 0; i < 5; i++) sc += (W1[i] * W2[i]);
  return sc;
}
bool checking(int &can, int &n, vector<vector<int> > &V) {
  for (int i = 0; i < n; i++)
    if (i != can)
      for (int k = i + 1; k < n; k++)
        if (k != can) {
          vector<vector<int> > W(2, vector<int>(5));
          for (int j = 0; j < 5; j++) {
            W[0][j] = V[can][j] - V[i][j];
            W[1][j] = V[can][j] - V[k][j];
          }
          if (scalar(W[0], W[1]) > 0) return false;
        }
  return true;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<vector<int> > V(n, vector<int>(5));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", &V[i][j]);
  if (n == 1)
    printf("1\n1\n");
  else if (n == 2)
    printf("2\n1\n2\n");
  else {
    vector<bool> Imp(n, false);
    int can, sc1, sc2, sc3;
    for (int i = 0; i + 1 < n; i++) {
      sc1 = sc2 = sc3 = 0, can = (i + 2) % n;
      vector<vector<int> > W(6, vector<int>(5));
      for (int j = 0; j < 5; j++) {
        W[0][j] = V[can][j] - V[i][j];
        W[1][j] = V[can][j] - V[i + 1][j];
        W[2][j] = V[i][j] - V[can][j];
        W[3][j] = V[i][j] - V[i + 1][j];
        W[4][j] = V[i + 1][j] - V[can][j];
        W[5][j] = V[i + 1][j] - V[i][j];
      }
      if (scalar(W[0], W[1]) > 0) Imp[can] = true;
      if (scalar(W[2], W[3]) > 0) Imp[i] = true;
      if (scalar(W[4], W[5]) > 0) Imp[i + 1] = true;
    }
    vector<int> Good;
    for (int i = 0; i < n; i++)
      if (!Imp[i] && checking(i, n, V)) Good.push_back(i);
    printf("%d\n", Good.size());
    for (int i = 0; i < Good.size(); i++) printf("%d\n", Good[i] + 1);
  }
  return 0;
}

### 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;
int scalar(vector<int> &W1, vector<int> &W2) {
  int sc = 0;
  for (int i = 0; i < 5; i++) sc += (W1[i] * W2[i]);
  return sc;
}
bool checking(int &can, int &n, vector<vector<int> > &V) {
  for (int i = 0; i < n; i++)
    if (i != can)
      for (int k = i + 1; k < n; k++)
        if (k != can) {
          vector<vector<int> > W(2, vector<int>(5));
          for (int j = 0; j < 5; j++) {
            W[0][j] = V[can][j] - V[i][j];
            W[1][j] = V[can][j] - V[k][j];
          }
          if (scalar(W[0], W[1]) > 0) return false;
        }
  return true;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<vector<int> > V(n, vector<int>(5));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 5; j++) scanf("%d", &V[i][j]);
  if (n == 1)
    printf("1\n1\n");
  else if (n == 2)
    printf("2\n1\n2\n");
  else {
    vector<bool> Imp(n, false);
    int can, sc1, sc2, sc3;
    for (int i = 0; i + 1 < n; i++) {
      sc1 = sc2 = sc3 = 0, can = (i + 2) % n;
      vector<vector<int> > W(6, vector<int>(5));
      for (int j = 0; j < 5; j++) {
        W[0][j] = V[can][j] - V[i][j];
        W[1][j] = V[can][j] - V[i + 1][j];
        W[2][j] = V[i][j] - V[can][j];
        W[3][j] = V[i][j] - V[i + 1][j];
        W[4][j] = V[i + 1][j] - V[can][j];
        W[5][j] = V[i + 1][j] - V[i][j];
      }
      if (scalar(W[0], W[1]) > 0) Imp[can] = true;
      if (scalar(W[2], W[3]) > 0) Imp[i] = true;
      if (scalar(W[4], W[5]) > 0) Imp[i + 1] = true;
    }
    vector<int> Good;
    for (int i = 0; i < n; i++)
      if (!Imp[i] && checking(i, n, V)) Good.push_back(i);
    printf("%d\n", Good.size());
    for (int i = 0; i < Good.size(); i++) printf("%d\n", Good[i] + 1);
  }
  return 0;
}

```