#include <bits/stdc++.h>
using namespace std;
struct point {
  int a, b, c, d, e;
};
point A[1001], B[1001][1001];
int calc(point A, point B) {
  return (A.a * B.a + A.b * B.b + A.c * B.c + A.d * B.d + A.e * B.e);
}
int n, sl = 0;
bool res[1001];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) res[i] = false;
  for (int i = 1; i <= n; i++)
    cin >> A[i].a >> A[i].b >> A[i].c >> A[i].d >> A[i].e;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j) {
        B[i][j].a = A[i].a - A[j].a;
        B[i][j].b = A[i].b - A[j].b;
        B[i][j].c = A[i].c - A[j].c;
        B[i][j].d = A[i].d - A[j].d;
        B[i][j].e = A[i].e - A[j].e;
      }
  for (int i = 1; i <= n; i++)
    if (!res[i]) {
      for (int j = 1; j <= n; j++)
        if (i != j) {
          for (int k = 1; k <= n; k++)
            if (k != j && k != i) {
              if (calc(B[i][j], B[i][k]) > 0) res[i] = true;
              if (calc(B[j][i], B[j][k]) > 0) res[j] = true;
              if (calc(B[k][j], B[k][i]) > 0) res[k] = true;
            }
        }
    }
  for (int i = 1; i <= n; i++)
    if (!res[i]) sl++;
  cout << sl << endl;
  for (int i = 1; i <= n; i++)
    if (!res[i]) cout << i << endl;
  return 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;
struct point {
  int a, b, c, d, e;
};
point A[1001], B[1001][1001];
int calc(point A, point B) {
  return (A.a * B.a + A.b * B.b + A.c * B.c + A.d * B.d + A.e * B.e);
}
int n, sl = 0;
bool res[1001];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) res[i] = false;
  for (int i = 1; i <= n; i++)
    cin >> A[i].a >> A[i].b >> A[i].c >> A[i].d >> A[i].e;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j) {
        B[i][j].a = A[i].a - A[j].a;
        B[i][j].b = A[i].b - A[j].b;
        B[i][j].c = A[i].c - A[j].c;
        B[i][j].d = A[i].d - A[j].d;
        B[i][j].e = A[i].e - A[j].e;
      }
  for (int i = 1; i <= n; i++)
    if (!res[i]) {
      for (int j = 1; j <= n; j++)
        if (i != j) {
          for (int k = 1; k <= n; k++)
            if (k != j && k != i) {
              if (calc(B[i][j], B[i][k]) > 0) res[i] = true;
              if (calc(B[j][i], B[j][k]) > 0) res[j] = true;
              if (calc(B[k][j], B[k][i]) > 0) res[k] = true;
            }
        }
    }
  for (int i = 1; i <= n; i++)
    if (!res[i]) sl++;
  cout << sl << endl;
  for (int i = 1; i <= n; i++)
    if (!res[i]) cout << i << endl;
  return 0;
}

```