#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const double eps = 1e-9;
const int MAXN = 100 + 23;
struct Point {
  double first[6];
  Point() {}
  Point(double x1, double x2, double x3, double x4, double x5) {
    first[0] = x1;
    first[1] = x2;
    first[2] = x3;
    first[3] = x4;
    first[4] = x5;
  }
  Point operator-(Point a) {
    return Point(first[0] - a.first[0], first[1] - a.first[1],
                 first[2] - a.first[2], first[3] - a.first[3],
                 first[4] - a.first[4]);
  }
} b[1000 + 3];
int a[1000 + 3];
int n;
double dot(Point first, Point second) {
  double res = 0;
  for (int i = 0; i < 5; i++) res += first.first[i] * second.first[i];
  return res;
}
double dist(Point first, Point second) {
  double res;
  for (int i = 0; i < 5; i++)
    res +=
        (first.first[i] - second.first[i]) * (first.first[i] - second.first[i]);
  return sqrt(res);
}
double ang(int first, int second, int z) {
  return acos(dot(b[second] - b[first], b[z] - b[first]) /
              (dist(b[second], b[first]) * dist(b[z], b[first])));
}
int main() {
  while (scanf("%d", &n) != -1) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 5; j++) scanf("%lf", &b[i].first[j]);
    }
    if (n > 100) {
      puts("0");
      continue;
    }
    memset(a, 0, sizeof(a));
    int ans = 0;
    for (int i = 0; i < n; i++) {
      int flag = 0;
      for (int j = 0; j < n; j++) {
        if (i == j) continue;
        for (int k = 0; k < n; k++) {
          if (k == i || k == j) continue;
          if (dot(b[j] - b[i], b[k] - b[i]) > eps) {
            flag = 1;
            a[i]++;
          }
        }
      }
      if (flag) ans++;
    }
    printf("%d\n", n - ans);
    for (int i = 0; i < n; i++) {
      if (!a[i]) {
        printf("%d\n", i + 1);
      }
    }
  }
  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;
const long long MOD = 1e9 + 7;
const double eps = 1e-9;
const int MAXN = 100 + 23;
struct Point {
  double first[6];
  Point() {}
  Point(double x1, double x2, double x3, double x4, double x5) {
    first[0] = x1;
    first[1] = x2;
    first[2] = x3;
    first[3] = x4;
    first[4] = x5;
  }
  Point operator-(Point a) {
    return Point(first[0] - a.first[0], first[1] - a.first[1],
                 first[2] - a.first[2], first[3] - a.first[3],
                 first[4] - a.first[4]);
  }
} b[1000 + 3];
int a[1000 + 3];
int n;
double dot(Point first, Point second) {
  double res = 0;
  for (int i = 0; i < 5; i++) res += first.first[i] * second.first[i];
  return res;
}
double dist(Point first, Point second) {
  double res;
  for (int i = 0; i < 5; i++)
    res +=
        (first.first[i] - second.first[i]) * (first.first[i] - second.first[i]);
  return sqrt(res);
}
double ang(int first, int second, int z) {
  return acos(dot(b[second] - b[first], b[z] - b[first]) /
              (dist(b[second], b[first]) * dist(b[z], b[first])));
}
int main() {
  while (scanf("%d", &n) != -1) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 5; j++) scanf("%lf", &b[i].first[j]);
    }
    if (n > 100) {
      puts("0");
      continue;
    }
    memset(a, 0, sizeof(a));
    int ans = 0;
    for (int i = 0; i < n; i++) {
      int flag = 0;
      for (int j = 0; j < n; j++) {
        if (i == j) continue;
        for (int k = 0; k < n; k++) {
          if (k == i || k == j) continue;
          if (dot(b[j] - b[i], b[k] - b[i]) > eps) {
            flag = 1;
            a[i]++;
          }
        }
      }
      if (flag) ans++;
    }
    printf("%d\n", n - ans);
    for (int i = 0; i < n; i++) {
      if (!a[i]) {
        printf("%d\n", i + 1);
      }
    }
  }
  return 0;
}

```