#include <bits/stdc++.h>
using namespace std;
long long r, d, n;
vector<vector<long long>> data;
const long double pi = 3.1415926535897932;
long long per(vector<long long> x, vector<long long> y) {
  long long su = 0;
  for (long long(i) = 0; (i) < (5); (i)++) {
    su += x[i] * y[i];
  }
  return su;
}
long double pere(vector<long long> x, vector<long long> y) {
  long double su = per(x, y);
  su /= sqrt(per(x, x)) * sqrt(per(y, y));
  su = acos(su);
  return su;
}
int main() {
  cin >> n;
  data.resize(n);
  for (long long(i) = 0; (i) < (n); (i)++) {
    for (long long(j) = 0; (j) < (5); (j)++) {
      data[i].resize(5);
      cin >> data[i][j];
    }
  }
  if (n > 32) {
    cout << 0;
    return 0;
  }
  vector<long long> result;
  for (long long(i) = 0; (i) < (n); (i)++) {
    bool mark = false;
    for (long long(j) = 0; (j) < (n); (j)++) {
      if (mark) {
        break;
      }
      for (long long(l) = 0; (l) < (n); (l)++) {
        if (i == j || j == l || i == l) {
          continue;
        }
        vector<long long> x, y;
        x.resize(5);
        y.resize(5);
        for (long long(h) = 0; (h) < (5); (h)++) {
          x[h] = data[j][h] - data[i][h];
          y[h] = data[l][h] - data[i][h];
        }
        if (pere(x, y) < pi / 2) {
          mark = true;
          break;
        }
      }
    }
    if (!mark) {
      result.push_back(i);
    }
  }
  cout << result.size() << endl;
  for (long long(i) = 0; (i) < (result.size()); (i)++) {
    cout << result[i] + 1 << endl;
  }
}

### Prompt

Create a solution in cpp for 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;
long long r, d, n;
vector<vector<long long>> data;
const long double pi = 3.1415926535897932;
long long per(vector<long long> x, vector<long long> y) {
  long long su = 0;
  for (long long(i) = 0; (i) < (5); (i)++) {
    su += x[i] * y[i];
  }
  return su;
}
long double pere(vector<long long> x, vector<long long> y) {
  long double su = per(x, y);
  su /= sqrt(per(x, x)) * sqrt(per(y, y));
  su = acos(su);
  return su;
}
int main() {
  cin >> n;
  data.resize(n);
  for (long long(i) = 0; (i) < (n); (i)++) {
    for (long long(j) = 0; (j) < (5); (j)++) {
      data[i].resize(5);
      cin >> data[i][j];
    }
  }
  if (n > 32) {
    cout << 0;
    return 0;
  }
  vector<long long> result;
  for (long long(i) = 0; (i) < (n); (i)++) {
    bool mark = false;
    for (long long(j) = 0; (j) < (n); (j)++) {
      if (mark) {
        break;
      }
      for (long long(l) = 0; (l) < (n); (l)++) {
        if (i == j || j == l || i == l) {
          continue;
        }
        vector<long long> x, y;
        x.resize(5);
        y.resize(5);
        for (long long(h) = 0; (h) < (5); (h)++) {
          x[h] = data[j][h] - data[i][h];
          y[h] = data[l][h] - data[i][h];
        }
        if (pere(x, y) < pi / 2) {
          mark = true;
          break;
        }
      }
    }
    if (!mark) {
      result.push_back(i);
    }
  }
  cout << result.size() << endl;
  for (long long(i) = 0; (i) < (result.size()); (i)++) {
    cout << result[i] + 1 << endl;
  }
}

```