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.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 100, maxm = 1e6 + 100, inf = 0x3f3f3f3f, mod = 1e9 + 7;
const long long INF = 1e18 + 7;
bool vis[maxn];
int point[maxn][10];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 5; j++) scanf("%d", &point[i][j]);
  }
  if (n == 1)
    printf("1\n1\n");
  else if (n == 2)
    printf("2\n1 2\n");
  else {
    memset(vis, false, sizeof(vis));
    bool flag = false;
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      if (vis[i]) continue;
      bool sign = false;
      for (int j = 1; j <= n; j++) {
        if (j == i) continue;
        for (int t = 1; t <= n; t++) {
          if (t == i || t == j) continue;
          int cou = 0;
          for (int k = 1; k <= 5; k++)
            cou += (point[j][k] - point[i][k]) * (point[t][k] - point[i][k]);
          if (cou > 0)
            sign = true;
          else if (cou < 0)
            vis[j] = vis[t] = true;
          if (sign) break;
        }
        if (sign) break;
      }
      if (!sign) flag = true, ans = i;
      if (flag) break;
    }
    if (flag)
      printf("1\n%d\n", ans);
    else
      printf("0\n");
  }
  return 0;
}
