#include <bits/stdc++.h>
using namespace std;
const long long mod1 = 1e12 + 193;
const long long mod2 = 1e16 + 371;
const long long mod = 1e9 + 7;
double dot(vector<int> x, vector<int> y) {
  double ret = 0;
  for (int i = 0; i < 5; i++) ret += x[i] * y[i];
  return ret;
}
double len(vector<int> x) { return sqrt(dot(x, x)); }
int main() {
  int n;
  cin >> n;
  if (n == 1) {
    cout << "1\n1";
    return 0;
  }
  if (n == 2) {
    cout << "2\n1\n2";
    return 0;
  }
  if (n > 11) {
    cout << 0;
    return 0;
  }
  vector<vector<int>> v(n, vector<int>(5));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) cin >> v[i][j];
  }
  for (int a = 0; a < n; a++) {
    bool ok = 1;
    for (int b = 0; b < n; b++) {
      for (int c = 0; c < n; c++) {
        if (a == b || a == c || b == c) continue;
        vector<int> x(5), y(5);
        for (int i = 0; i < 5; i++)
          x[i] = v[b][i] - v[a][i], y[i] = v[c][i] - v[a][i];
        double ang = acos(dot(x, y) / (len(x) * len(y))) * 180.0 / acos(-1);
        if (90.0 - ang >= 1e-8) ok = 0;
      }
    }
    if (ok) {
      cout << "1\n" << a + 1;
      return 0;
    }
  }
  cout << 0;
}

### Prompt

Please formulate a cpp solution to 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 long long mod1 = 1e12 + 193;
const long long mod2 = 1e16 + 371;
const long long mod = 1e9 + 7;
double dot(vector<int> x, vector<int> y) {
  double ret = 0;
  for (int i = 0; i < 5; i++) ret += x[i] * y[i];
  return ret;
}
double len(vector<int> x) { return sqrt(dot(x, x)); }
int main() {
  int n;
  cin >> n;
  if (n == 1) {
    cout << "1\n1";
    return 0;
  }
  if (n == 2) {
    cout << "2\n1\n2";
    return 0;
  }
  if (n > 11) {
    cout << 0;
    return 0;
  }
  vector<vector<int>> v(n, vector<int>(5));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) cin >> v[i][j];
  }
  for (int a = 0; a < n; a++) {
    bool ok = 1;
    for (int b = 0; b < n; b++) {
      for (int c = 0; c < n; c++) {
        if (a == b || a == c || b == c) continue;
        vector<int> x(5), y(5);
        for (int i = 0; i < 5; i++)
          x[i] = v[b][i] - v[a][i], y[i] = v[c][i] - v[a][i];
        double ang = acos(dot(x, y) / (len(x) * len(y))) * 180.0 / acos(-1);
        if (90.0 - ang >= 1e-8) ok = 0;
      }
    }
    if (ok) {
      cout << "1\n" << a + 1;
      return 0;
    }
  }
  cout << 0;
}

```