#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000 + 10;
const double PI = acos(-1.0);
struct Point {
  double data[5];
} points[MAXN];
queue<int> q;
double across(Point a, Point b, Point c) {
  double v1[5], v2[5];
  for (int i = 0; i < 5; i++) {
    v1[i] = b.data[i] - a.data[i];
    v2[i] = c.data[i] - a.data[i];
  }
  double numerator = 0;
  for (int i = 0; i < 5; i++) {
    numerator += v1[i] * v2[i];
  }
  double denominator = 0, sum1 = 0, sum2 = 0;
  for (int i = 0; i < 5; i++) {
    sum1 += v1[i] * v1[i];
    sum2 += v2[i] * v2[i];
  }
  denominator = sqrt(sum1) * sqrt(sum2);
  return 1.0 * numerator / denominator;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%lf", &points[i].data[j]);
    }
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    bool OK = true;
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        if (across(points[i], points[j], points[k]) > 0) {
          OK = false;
          break;
        }
      }
      if (!OK) break;
    }
    if (OK) {
      cnt++;
      q.push(i + 1);
    }
  }
  printf("%d\n", cnt);
  while (!q.empty()) {
    printf("%d\n", q.front());
    q.pop();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const int MAXN = 1000 + 10;
const double PI = acos(-1.0);
struct Point {
  double data[5];
} points[MAXN];
queue<int> q;
double across(Point a, Point b, Point c) {
  double v1[5], v2[5];
  for (int i = 0; i < 5; i++) {
    v1[i] = b.data[i] - a.data[i];
    v2[i] = c.data[i] - a.data[i];
  }
  double numerator = 0;
  for (int i = 0; i < 5; i++) {
    numerator += v1[i] * v2[i];
  }
  double denominator = 0, sum1 = 0, sum2 = 0;
  for (int i = 0; i < 5; i++) {
    sum1 += v1[i] * v1[i];
    sum2 += v2[i] * v2[i];
  }
  denominator = sqrt(sum1) * sqrt(sum2);
  return 1.0 * numerator / denominator;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%lf", &points[i].data[j]);
    }
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    bool OK = true;
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        if (across(points[i], points[j], points[k]) > 0) {
          OK = false;
          break;
        }
      }
      if (!OK) break;
    }
    if (OK) {
      cnt++;
      q.push(i + 1);
    }
  }
  printf("%d\n", cnt);
  while (!q.empty()) {
    printf("%d\n", q.front());
    q.pop();
  }
  return 0;
}

```