#include <bits/stdc++.h>
using namespace std;
int tot;
int num[1008];
int is[1005];
int n, pos[1005][5];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", pos[i] + j);
    }
  }
  int flag = 0;
  int yes = 0;
  if (n < 3) {
    if (n == 1) {
      printf("1\n1\n");
    } else {
      printf("2\n1 2\n");
    }
    return 0;
  }
  for (int i = 0; i < n; i++) {
    flag = 1;
    if (is[i]) continue;
    is[i] = 1;
    for (int j = 0; j < n && flag; j++) {
      for (int k = 0; k < n && flag; k++) {
        if (i == j || i == k || k == j) {
          continue;
        }
        int x[5] = {0}, y[5] = {0};
        for (int m = 0; m < 5; m++) {
          x[m] = pos[j][m] - pos[i][m];
        }
        for (int m = 0; m < 5; m++) {
          y[m] = pos[k][m] - pos[i][m];
        }
        int temp = 0;
        for (int m = 0; m < 5; m++) {
          temp += x[m] * y[m];
        }
        if (temp <= 0) {
          is[j] = 1;
          is[k] = 1;
        } else {
          is[i] = 1;
          flag = 0;
        }
      }
    }
    if (flag == 1) {
      printf("1\n%d", i + 1);
      yes = 1;
      break;
    }
  }
  if (yes == 0) {
    printf("0\n");
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
int tot;
int num[1008];
int is[1005];
int n, pos[1005][5];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 5; j++) {
      scanf("%d", pos[i] + j);
    }
  }
  int flag = 0;
  int yes = 0;
  if (n < 3) {
    if (n == 1) {
      printf("1\n1\n");
    } else {
      printf("2\n1 2\n");
    }
    return 0;
  }
  for (int i = 0; i < n; i++) {
    flag = 1;
    if (is[i]) continue;
    is[i] = 1;
    for (int j = 0; j < n && flag; j++) {
      for (int k = 0; k < n && flag; k++) {
        if (i == j || i == k || k == j) {
          continue;
        }
        int x[5] = {0}, y[5] = {0};
        for (int m = 0; m < 5; m++) {
          x[m] = pos[j][m] - pos[i][m];
        }
        for (int m = 0; m < 5; m++) {
          y[m] = pos[k][m] - pos[i][m];
        }
        int temp = 0;
        for (int m = 0; m < 5; m++) {
          temp += x[m] * y[m];
        }
        if (temp <= 0) {
          is[j] = 1;
          is[k] = 1;
        } else {
          is[i] = 1;
          flag = 0;
        }
      }
    }
    if (flag == 1) {
      printf("1\n%d", i + 1);
      yes = 1;
      break;
    }
  }
  if (yes == 0) {
    printf("0\n");
  }
  return 0;
}

```