#include <bits/stdc++.h>
using namespace std;
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
template <class T>
inline T bigmod(T p, T e, T M) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    long long int t = bigmod(p, e / 2, M);
    return (T)((t * t) % M);
  }
  return (T)((long long int)bigmod(p, e - 1, M) * (long long int)p) % M;
}
template <class T>
inline T bigexp(T p, T e) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    long long int t = bigexp(p, e / 2);
    return (T)((t * t));
  }
  return (T)((long long int)bigexp(p, e - 1) * (long long int)p);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
int dx4[] = {1, 0, -1, 0};
int dy4[] = {0, 1, 0, -1};
int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy8[] = {0, 1, 1, 1, 0, -1, -1, -1};
int nx8[] = {1, 1, -1, -1, 2, 2, -2, -2};
int ny8[] = {2, -2, 2, -2, 1, -1, 1, -1};
int month[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int point[1005][5], n, ans[1005], ab[5], ac[5];
int call(int i, int j, int k) {
  for (__typeof(4) p = (0); p <= (4); p++) ab[p] = point[j][p] - point[i][p];
  for (__typeof(4) p = (0); p <= (4); p++) ac[p] = point[k][p] - point[i][p];
  double abdac = 0;
  for (__typeof(4) p = (0); p <= (4); p++) abdac += (ab[p] * ac[p]);
  double aba = (double)sqrt((double)ab[0] * ab[0] + ab[1] * ab[1] +
                            ab[2] * ab[2] + ab[3] * ab[3] + ab[4] * ab[4]);
  aba *= (double)sqrt((double)ac[0] * ac[0] + ac[1] * ac[1] + ac[2] * ac[2] +
                      ac[3] * ac[3] + ac[4] * ac[4]);
  double tmp = abdac / aba;
  tmp = acos((double)tmp) * ((double)180.0 / 3.14159);
  if (tmp >= 90.0)
    return 0;
  else
    return 1;
}
int main() {
  scanf("%d", &n);
  for (__typeof(n) i = (1); i <= (n); i++) {
    scanf("%d %d", &point[i][0], &point[i][1]);
    scanf("%d %d", &point[i][2], &point[i][3]);
    scanf("%d", &point[i][4]);
  }
  int cnt = 0;
  int res = 0;
  for (__typeof(n) i = (1); i <= (n); i++) {
    cnt = 0;
    for (__typeof(n) j = (1); j <= (n); j++) {
      if (i != j) {
        for (__typeof(n) k = (1); k <= (n); k++) {
          if (i != k && j != k) {
            cnt += call(i, j, k);
            if (cnt == 1) break;
          }
        }
      }
      if (cnt == 1) break;
    }
    if (cnt == 0) {
      ans[++res] = i;
    }
  }
  cout << res, printf("\n");
  for (__typeof(res) i = (1); i <= (res); i++) printf("%d ", ans[i]);
  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;
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
template <class T>
inline T bigmod(T p, T e, T M) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    long long int t = bigmod(p, e / 2, M);
    return (T)((t * t) % M);
  }
  return (T)((long long int)bigmod(p, e - 1, M) * (long long int)p) % M;
}
template <class T>
inline T bigexp(T p, T e) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    long long int t = bigexp(p, e / 2);
    return (T)((t * t));
  }
  return (T)((long long int)bigexp(p, e - 1) * (long long int)p);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
int dx4[] = {1, 0, -1, 0};
int dy4[] = {0, 1, 0, -1};
int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy8[] = {0, 1, 1, 1, 0, -1, -1, -1};
int nx8[] = {1, 1, -1, -1, 2, 2, -2, -2};
int ny8[] = {2, -2, 2, -2, 1, -1, 1, -1};
int month[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int point[1005][5], n, ans[1005], ab[5], ac[5];
int call(int i, int j, int k) {
  for (__typeof(4) p = (0); p <= (4); p++) ab[p] = point[j][p] - point[i][p];
  for (__typeof(4) p = (0); p <= (4); p++) ac[p] = point[k][p] - point[i][p];
  double abdac = 0;
  for (__typeof(4) p = (0); p <= (4); p++) abdac += (ab[p] * ac[p]);
  double aba = (double)sqrt((double)ab[0] * ab[0] + ab[1] * ab[1] +
                            ab[2] * ab[2] + ab[3] * ab[3] + ab[4] * ab[4]);
  aba *= (double)sqrt((double)ac[0] * ac[0] + ac[1] * ac[1] + ac[2] * ac[2] +
                      ac[3] * ac[3] + ac[4] * ac[4]);
  double tmp = abdac / aba;
  tmp = acos((double)tmp) * ((double)180.0 / 3.14159);
  if (tmp >= 90.0)
    return 0;
  else
    return 1;
}
int main() {
  scanf("%d", &n);
  for (__typeof(n) i = (1); i <= (n); i++) {
    scanf("%d %d", &point[i][0], &point[i][1]);
    scanf("%d %d", &point[i][2], &point[i][3]);
    scanf("%d", &point[i][4]);
  }
  int cnt = 0;
  int res = 0;
  for (__typeof(n) i = (1); i <= (n); i++) {
    cnt = 0;
    for (__typeof(n) j = (1); j <= (n); j++) {
      if (i != j) {
        for (__typeof(n) k = (1); k <= (n); k++) {
          if (i != k && j != k) {
            cnt += call(i, j, k);
            if (cnt == 1) break;
          }
        }
      }
      if (cnt == 1) break;
    }
    if (cnt == 0) {
      ans[++res] = i;
    }
  }
  cout << res, printf("\n");
  for (__typeof(res) i = (1); i <= (res); i++) printf("%d ", ans[i]);
  return 0;
}

```