#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-10;
const long long MOD = 1e9 + 7;
const long long OO = 1e18 + 9;
const int N = 1e3 + 3;
struct Point {
  int p[5];
  int operator*(const Point &other) const {
    int ret = 0;
    for (int i = 0; i < 5; ++i) ret += p[i] * other.p[i];
    return ret;
  }
  Point operator-(const Point &other) const {
    Point np;
    for (int i = 0; i < 5; ++i) np.p[i] = p[i] - other.p[i];
    return np;
  }
};
int n;
vector<Point> points;
vector<Point> getVectors(int idx) {
  vector<Point> ret;
  for (int i = 0; i < n; ++i) {
    if (i == idx) continue;
    ret.push_back(points[i] - points[idx]);
  }
  return ret;
}
bool good(int idx) {
  vector<Point> vectors = getVectors(idx);
  if (vectors.empty()) return true;
  int tr = 1e4;
  while (tr--) {
    int r1 = rand() % vectors.size();
    int r2 = rand() % vectors.size();
    if (r1 == r2) continue;
    if (vectors[r1] * vectors[r2] > 0) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  srand(time(0));
  cin >> n;
  points.resize(n);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 5; ++j) cin >> points[i].p[j];
  }
  vector<int> ans;
  for (int i = 0; i < n; ++i)
    if (good(i)) ans.push_back(i + 1);
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); ++i) cout << ans[i] << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 double EPS = 1e-10;
const long long MOD = 1e9 + 7;
const long long OO = 1e18 + 9;
const int N = 1e3 + 3;
struct Point {
  int p[5];
  int operator*(const Point &other) const {
    int ret = 0;
    for (int i = 0; i < 5; ++i) ret += p[i] * other.p[i];
    return ret;
  }
  Point operator-(const Point &other) const {
    Point np;
    for (int i = 0; i < 5; ++i) np.p[i] = p[i] - other.p[i];
    return np;
  }
};
int n;
vector<Point> points;
vector<Point> getVectors(int idx) {
  vector<Point> ret;
  for (int i = 0; i < n; ++i) {
    if (i == idx) continue;
    ret.push_back(points[i] - points[idx]);
  }
  return ret;
}
bool good(int idx) {
  vector<Point> vectors = getVectors(idx);
  if (vectors.empty()) return true;
  int tr = 1e4;
  while (tr--) {
    int r1 = rand() % vectors.size();
    int r2 = rand() % vectors.size();
    if (r1 == r2) continue;
    if (vectors[r1] * vectors[r2] > 0) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  srand(time(0));
  cin >> n;
  points.resize(n);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 5; ++j) cin >> points[i].p[j];
  }
  vector<int> ans;
  for (int i = 0; i < n; ++i)
    if (good(i)) ans.push_back(i + 1);
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); ++i) cout << ans[i] << endl;
}

```