#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
struct Point {
  int id;
  double a, b, c, d, e;
  double operator*(const struct Point& aa) const {
    double xy = aa.a * a + aa.b * b + aa.c * c + aa.d * d + aa.e * e;
    double len1 = sqrt(aa.a * aa.a + aa.b * aa.b + aa.c * aa.c + aa.d * aa.d +
                       aa.e * aa.e);
    double len2 = sqrt(a * a + b * b + c * c + d * d + e * e);
    double ang = acos(xy / (len1 * len2)) * 180.0 / pi;
    return ang;
  }
  struct Point operator-() const {
    struct Point aa;
    aa.a = -a, aa.b = -b, aa.c = -c, aa.d = -d, aa.e = -e;
    return aa;
  }
} p[1005], p1[1005];
int vis[1005], n;
int cnt, ans[1005];
struct Point getPoint(int i, int j) {
  struct Point aa;
  aa.a = p[j].a - p[i].a;
  aa.b = p[j].b - p[i].b;
  aa.c = p[j].c - p[i].c;
  aa.d = p[j].d - p[i].d;
  aa.e = p[j].e - p[i].e;
  return aa;
};
int main() {
  cnt = 0;
  memset(vis, 0, sizeof(vis));
  cin >> n;
  for (int i = 1; i <= n; i++)
    scanf("%lf%lf%lf%lf%lf", &p[i].a, &p[i].b, &p[i].c, &p[i].d, &p[i].e);
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      int num = 0;
      for (int j = 1; j <= n; j++) {
        if (j == i) continue;
        p1[num] = getPoint(i, j);
        p1[num].id = j;
        num++;
      }
      int flag = 1;
      for (int j = 0; j < num - 1 && flag; j++) {
        for (int k = j + 1; k < num; k++) {
          double ang = p1[j] * p1[k];
          if (ang < 90.0) {
            flag = 0;
          }
          struct Point bc = getPoint(p1[j].id, p1[k].id);
          if (bc * (-p1[j]) < 90.0) {
            vis[p1[j].id] = 1;
          }
          if ((-bc) * (-p1[k]) < 90.0) {
            vis[p1[k].id] = 1;
          }
        }
      }
      if (flag) ans[cnt++] = i;
    }
  }
  cout << cnt << endl;
  for (int i = 0; i < cnt; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Please create a solution in CPP 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 double pi = acos(-1.0);
struct Point {
  int id;
  double a, b, c, d, e;
  double operator*(const struct Point& aa) const {
    double xy = aa.a * a + aa.b * b + aa.c * c + aa.d * d + aa.e * e;
    double len1 = sqrt(aa.a * aa.a + aa.b * aa.b + aa.c * aa.c + aa.d * aa.d +
                       aa.e * aa.e);
    double len2 = sqrt(a * a + b * b + c * c + d * d + e * e);
    double ang = acos(xy / (len1 * len2)) * 180.0 / pi;
    return ang;
  }
  struct Point operator-() const {
    struct Point aa;
    aa.a = -a, aa.b = -b, aa.c = -c, aa.d = -d, aa.e = -e;
    return aa;
  }
} p[1005], p1[1005];
int vis[1005], n;
int cnt, ans[1005];
struct Point getPoint(int i, int j) {
  struct Point aa;
  aa.a = p[j].a - p[i].a;
  aa.b = p[j].b - p[i].b;
  aa.c = p[j].c - p[i].c;
  aa.d = p[j].d - p[i].d;
  aa.e = p[j].e - p[i].e;
  return aa;
};
int main() {
  cnt = 0;
  memset(vis, 0, sizeof(vis));
  cin >> n;
  for (int i = 1; i <= n; i++)
    scanf("%lf%lf%lf%lf%lf", &p[i].a, &p[i].b, &p[i].c, &p[i].d, &p[i].e);
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      int num = 0;
      for (int j = 1; j <= n; j++) {
        if (j == i) continue;
        p1[num] = getPoint(i, j);
        p1[num].id = j;
        num++;
      }
      int flag = 1;
      for (int j = 0; j < num - 1 && flag; j++) {
        for (int k = j + 1; k < num; k++) {
          double ang = p1[j] * p1[k];
          if (ang < 90.0) {
            flag = 0;
          }
          struct Point bc = getPoint(p1[j].id, p1[k].id);
          if (bc * (-p1[j]) < 90.0) {
            vis[p1[j].id] = 1;
          }
          if ((-bc) * (-p1[k]) < 90.0) {
            vis[p1[k].id] = 1;
          }
        }
      }
      if (flag) ans[cnt++] = i;
    }
  }
  cout << cnt << endl;
  for (int i = 0; i < cnt; i++) printf("%d\n", ans[i]);
  return 0;
}

```