The Little Elephant has two permutations a and b of length n, consisting of numbers from 1 to n, inclusive. Let's denote the i-th (1 ≤ i ≤ n) element of the permutation a as ai, the j-th (1 ≤ j ≤ n) element of the permutation b — as bj.

The distance between permutations a and b is the minimum absolute value of the difference between the positions of the occurrences of some number in a and in b. More formally, it's such minimum |i - j|, that ai = bj.

A cyclic shift number i (1 ≤ i ≤ n) of permutation b consisting from n elements is a permutation bibi + 1... bnb1b2... bi - 1. Overall a permutation has n cyclic shifts.

The Little Elephant wonders, for all cyclic shifts of permutation b, what is the distance between the cyclic shift and permutation a?

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the size of the permutations. The second line contains permutation a as n distinct numbers from 1 to n, inclusive. The numbers are separated with single spaces. The third line contains permutation b in the same format.

Output

In n lines print n integers — the answers for cyclic shifts. Print the answers to the shifts in the order of the shifts' numeration in permutation b, that is, first for the 1-st cyclic shift, then for the 2-nd, and so on.

Examples

Input

2
1 2
2 1


Output

1
0


Input

4
2 1 3 4
3 4 2 1


Output

2
1
0
1
#include <bits/stdc++.h>
using namespace std;
const int Maxn = 100 * 1000 + 10;
int n, inda[Maxn], indb[Maxn];
set<pair<int, int> > a, b, b1;
int main() {
  scanf("%d", &n);
  int aa;
  for (int i = 0; i < n; i++) {
    scanf("%d", &aa);
    aa--;
    inda[aa] = i;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &aa);
    aa--;
    indb[aa] = i;
  }
  for (int i = 0; i < n; i++) {
    if (indb[i] > inda[i])
      a.insert(make_pair(indb[i] - inda[i], inda[i]));
    else
      b.insert(make_pair(inda[i] - indb[i], indb[i])),
          b1.insert(make_pair(indb[i], inda[i] - indb[i]));
  }
  for (int i = 0; i < n; i++) {
    printf("%d\n",
           min((b.begin()->first + i) % n, (a.begin()->first - i + n) % n));
    if (b1.begin()->first == i) {
      int ind = b1.begin()->second + i;
      a.insert(make_pair(n - ind + i, ind));
      b.erase(make_pair(b1.begin()->second, b1.begin()->first));
      b1.erase(b1.begin());
    }
    while (!a.empty() && a.begin()->first == i + 1) {
      b.insert(make_pair(-i - 1, a.begin()->second + i + 1));
      b1.insert(make_pair(a.begin()->second + i + 1, -i - 1));
      a.erase(a.begin());
    }
  }
  return 0;
}
