#include <bits/stdc++.h>
using namespace std;
int quick_pow(int a, int b, int MOD) {
  a %= MOD;
  int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    b /= 2;
    a = (a * a) % MOD;
  }
  return res;
}
const int maxn = 200000 + 10;
const int MOD = 1e9 + 7;
int a[maxn], b[maxn], c[maxn], d[maxn], e[maxn], vis[maxn];
int main() {
  int i, j, n;
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    scanf("%d%d%d%d%d", &a[i], &b[i], &c[i], &d[i], &e[i]);
  if (n > 11) {
    puts("0");
    return 0;
  }
  int res = n;
  for (i = 1; i <= n; i++) {
    int yes = 0;
    for (j = 1; j <= n && !yes; j++) {
      if (j == i) continue;
      for (int k = 1; k <= n && !yes; k++) {
        if (k == i || k == j) continue;
        int A = a[j] - a[i];
        int B = b[j] - b[i];
        int C = c[j] - c[i];
        int D = d[j] - d[i];
        int E = e[j] - e[i];
        int AA = a[k] - a[i];
        int BB = b[k] - b[i];
        int CC = c[k] - c[i];
        int DD = d[k] - d[i];
        int EE = e[k] - e[i];
        double x = 1.0 * A * AA + B * BB + C * CC + D * DD + E * EE;
        double y = sqrt(1.0 * A * A + B * B + C * C + D * D + E * E) *
                   sqrt(1.0 * AA * AA + BB * BB + CC * CC + DD * DD + EE * EE);
        if (x / y > 0.0) {
          yes = 1;
          vis[i] = 1;
          res--;
          break;
        }
      }
    }
  }
  printf("%d\n", res);
  for (i = 1; i <= n; i++)
    if (vis[i] == 0) printf("%d ", i);
  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;
int quick_pow(int a, int b, int MOD) {
  a %= MOD;
  int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    b /= 2;
    a = (a * a) % MOD;
  }
  return res;
}
const int maxn = 200000 + 10;
const int MOD = 1e9 + 7;
int a[maxn], b[maxn], c[maxn], d[maxn], e[maxn], vis[maxn];
int main() {
  int i, j, n;
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    scanf("%d%d%d%d%d", &a[i], &b[i], &c[i], &d[i], &e[i]);
  if (n > 11) {
    puts("0");
    return 0;
  }
  int res = n;
  for (i = 1; i <= n; i++) {
    int yes = 0;
    for (j = 1; j <= n && !yes; j++) {
      if (j == i) continue;
      for (int k = 1; k <= n && !yes; k++) {
        if (k == i || k == j) continue;
        int A = a[j] - a[i];
        int B = b[j] - b[i];
        int C = c[j] - c[i];
        int D = d[j] - d[i];
        int E = e[j] - e[i];
        int AA = a[k] - a[i];
        int BB = b[k] - b[i];
        int CC = c[k] - c[i];
        int DD = d[k] - d[i];
        int EE = e[k] - e[i];
        double x = 1.0 * A * AA + B * BB + C * CC + D * DD + E * EE;
        double y = sqrt(1.0 * A * A + B * B + C * C + D * D + E * E) *
                   sqrt(1.0 * AA * AA + BB * BB + CC * CC + DD * DD + EE * EE);
        if (x / y > 0.0) {
          yes = 1;
          vis[i] = 1;
          res--;
          break;
        }
      }
    }
  }
  printf("%d\n", res);
  for (i = 1; i <= n; i++)
    if (vis[i] == 0) printf("%d ", i);
  return 0;
}

```