#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void chkmax(T& x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
inline void chkmin(T& x, U y) {
  if (y < x) x = y;
}
const int MX = 1333;
const int mod = 1000000007;
struct point {
  int a[5];
  void input() {
    for (int i = 0; i < 5; i++) scanf("%d", a + i);
  }
} p[MX];
point operator-(point lhs, point rhs) {
  point rlt;
  for (int i = 0; i < 5; i++) rlt.a[i] = lhs.a[i] - rhs.a[i];
  return rlt;
}
vector<int> ans;
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) p[i].input();
  if (n > 500) return puts("0"), 0;
  for (int i = 0; i < n; i++) {
    bool flag = 0;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      for (int k = j + 1; k < n; k++) {
        if (k == i) continue;
        long long cur = 0;
        point p1, p2;
        p1 = p[j] - p[i], p2 = p[k] - p[i];
        for (int t = 0; t < 5; t++) cur += 1LL * p1.a[t] * p2.a[t];
        if (cur > 0) flag = 1;
      }
    }
    if (!flag) ans.push_back(i + 1);
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    printf("%d\n", ans[i]);
  }
  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;
template <typename T, typename U>
inline void chkmax(T& x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
inline void chkmin(T& x, U y) {
  if (y < x) x = y;
}
const int MX = 1333;
const int mod = 1000000007;
struct point {
  int a[5];
  void input() {
    for (int i = 0; i < 5; i++) scanf("%d", a + i);
  }
} p[MX];
point operator-(point lhs, point rhs) {
  point rlt;
  for (int i = 0; i < 5; i++) rlt.a[i] = lhs.a[i] - rhs.a[i];
  return rlt;
}
vector<int> ans;
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) p[i].input();
  if (n > 500) return puts("0"), 0;
  for (int i = 0; i < n; i++) {
    bool flag = 0;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      for (int k = j + 1; k < n; k++) {
        if (k == i) continue;
        long long cur = 0;
        point p1, p2;
        p1 = p[j] - p[i], p2 = p[k] - p[i];
        for (int t = 0; t < 5; t++) cur += 1LL * p1.a[t] * p2.a[t];
        if (cur > 0) flag = 1;
      }
    }
    if (!flag) ans.push_back(i + 1);
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```