#include <bits/stdc++.h>
using namespace std;
struct node {
  int a, b, c, d, e;
};
int xy(node x, node y) {
  return (x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e);
}
int scalar(node x) {
  return x.a * x.a + x.b * x.b + x.c * x.c + x.d * x.d + x.e * x.e;
}
node vec(node x, node y) {
  node t;
  t.a = y.a - x.a;
  t.b = y.b - x.b;
  t.c = y.c - x.c;
  t.d = y.d - x.d;
  t.e = y.e - x.e;
  return t;
}
int gcd(int x, int y) {
  if (!y)
    return x;
  else
    return gcd(y, x % y);
}
node aa[1005];
int ans[1005];
int main() {
  int n, i, j, k;
  while (~scanf("%d", &n)) {
    for (i = 0; i < n; i++) {
      scanf("%d %d %d %d %d", &aa[i].a, &aa[i].b, &aa[i].c, &aa[i].d, &aa[i].e);
    }
    int top = 0;
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        if (j == i) continue;
        for (k = j + 1; k < n; k++) {
          if (k == i) continue;
          node x = vec(aa[i], aa[j]);
          node y = vec(aa[i], aa[k]);
          int t1 = xy(x, y);
          double t2 = 1.0 * sqrt(1.0 * scalar(x)) * sqrt(1.0 * scalar(y));
          double c = t1 / t2;
          if (c > 0) break;
        }
        if (k < n) break;
      }
      if (j == n) ans[top++] = i;
    }
    printf("%d\n", top);
    for (i = 0; i < top; i++) {
      if (i) printf(" ");
      printf("%d", ans[i] + 1);
    }
    if (top) printf("\n");
  }
  return 0;
}

### Prompt

Generate 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;
struct node {
  int a, b, c, d, e;
};
int xy(node x, node y) {
  return (x.a * y.a + x.b * y.b + x.c * y.c + x.d * y.d + x.e * y.e);
}
int scalar(node x) {
  return x.a * x.a + x.b * x.b + x.c * x.c + x.d * x.d + x.e * x.e;
}
node vec(node x, node y) {
  node t;
  t.a = y.a - x.a;
  t.b = y.b - x.b;
  t.c = y.c - x.c;
  t.d = y.d - x.d;
  t.e = y.e - x.e;
  return t;
}
int gcd(int x, int y) {
  if (!y)
    return x;
  else
    return gcd(y, x % y);
}
node aa[1005];
int ans[1005];
int main() {
  int n, i, j, k;
  while (~scanf("%d", &n)) {
    for (i = 0; i < n; i++) {
      scanf("%d %d %d %d %d", &aa[i].a, &aa[i].b, &aa[i].c, &aa[i].d, &aa[i].e);
    }
    int top = 0;
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        if (j == i) continue;
        for (k = j + 1; k < n; k++) {
          if (k == i) continue;
          node x = vec(aa[i], aa[j]);
          node y = vec(aa[i], aa[k]);
          int t1 = xy(x, y);
          double t2 = 1.0 * sqrt(1.0 * scalar(x)) * sqrt(1.0 * scalar(y));
          double c = t1 / t2;
          if (c > 0) break;
        }
        if (k < n) break;
      }
      if (j == n) ans[top++] = i;
    }
    printf("%d\n", top);
    for (i = 0; i < top; i++) {
      if (i) printf(" ");
      printf("%d", ans[i] + 1);
    }
    if (top) printf("\n");
  }
  return 0;
}

```