#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1005;
int n, tot;
int a[MAXN][6];
int g[MAXN];
int Read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x * 10) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
void Put1(int x) {
  if (x > 9) Put1(x / 10);
  putchar(x % 10 ^ 48);
}
void Put(int x, char c = -1) {
  if (x < 0) putchar('-'), x = -x;
  Put1(x);
  if (c >= 0) putchar(c);
}
template <typename T>
T Max(T x, T y) {
  return x > y ? x : y;
}
template <typename T>
T Min(T x, T y) {
  return x < y ? x : y;
}
template <typename T>
T Abs(T x) {
  return x < 0 ? -x : x;
}
int main() {
  n = Read();
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= 5; ++j) a[i][j] = Read();
  if (n > 11) {
    Put(0);
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    bool good = 1;
    for (int j = 1; j <= n && good; ++j) {
      if (i != j)
        for (int k = j + 1; k <= n && good; ++k)
          if (k != i) {
            int dz = 0;
            for (int l = 1; l <= 5; ++l)
              dz += (a[j][l] - a[i][l]) * (a[k][l] - a[i][l]);
            if (dz > 0) good = 0;
          }
    }
    if (good) g[++tot] = i;
  }
  Put(tot, '\n');
  for (int i = 1; i <= tot; ++i) Put(g[i], ' ');
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 = 1005;
int n, tot;
int a[MAXN][6];
int g[MAXN];
int Read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x * 10) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
void Put1(int x) {
  if (x > 9) Put1(x / 10);
  putchar(x % 10 ^ 48);
}
void Put(int x, char c = -1) {
  if (x < 0) putchar('-'), x = -x;
  Put1(x);
  if (c >= 0) putchar(c);
}
template <typename T>
T Max(T x, T y) {
  return x > y ? x : y;
}
template <typename T>
T Min(T x, T y) {
  return x < y ? x : y;
}
template <typename T>
T Abs(T x) {
  return x < 0 ? -x : x;
}
int main() {
  n = Read();
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= 5; ++j) a[i][j] = Read();
  if (n > 11) {
    Put(0);
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    bool good = 1;
    for (int j = 1; j <= n && good; ++j) {
      if (i != j)
        for (int k = j + 1; k <= n && good; ++k)
          if (k != i) {
            int dz = 0;
            for (int l = 1; l <= 5; ++l)
              dz += (a[j][l] - a[i][l]) * (a[k][l] - a[i][l]);
            if (dz > 0) good = 0;
          }
    }
    if (good) g[++tot] = i;
  }
  Put(tot, '\n');
  for (int i = 1; i <= tot; ++i) Put(g[i], ' ');
  return 0;
}

```