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;
multiset<int> S;
int a[100010], pos[100010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    pos[x] = i;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[i] = i - pos[x];
    S.insert(a[i]);
  }
  for (int i = 1; i <= n; i++) {
    multiset<int>::iterator k = S.lower_bound(i - 1);
    int ans = 1 << 30;
    if (k != S.end()) ans = min(ans, *k - (i - 1));
    if (k != S.begin()) ans = min(ans, (i - 1) - *--k);
    printf("%d\n", ans);
    S.erase(S.find(a[i]));
    S.insert(a[i] + n);
  }
  return (0);
}
