You are given two arrays a and b, both of length n. All elements of both arrays are from 0 to n-1.

You can reorder elements of the array b (if you want, you may leave the order of elements as it is). After that, let array c be the array of length n, the i-th element of this array is c_i = (a_i + b_i) \% n, where x \% y is x modulo y.

Your task is to reorder elements of the array b to obtain the lexicographically minimum possible array c.

Array x of length n is lexicographically less than array y of length n, if there exists such i (1 ≤ i ≤ n), that x_i < y_i, and for any j (1 ≤ j < i) x_j = y_j.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a, b and c.

The second line of the input contains n integers a_1, a_2, ..., a_n (0 ≤ a_i < n), where a_i is the i-th element of a.

The third line of the input contains n integers b_1, b_2, ..., b_n (0 ≤ b_i < n), where b_i is the i-th element of b.

Output

Print the lexicographically minimum possible array c. Recall that your task is to reorder elements of the array b and obtain the lexicographically minimum possible array c, where the i-th element of c is c_i = (a_i + b_i) \% n.

Examples

Input


4
0 1 2 1
3 2 1 1


Output


1 0 0 2 


Input


7
2 5 1 5 3 4 3
2 4 3 5 6 5 1


Output


0 0 0 1 0 2 4 
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
long long m, n, r, l, k;
long long flag, temp, cnt;
int a[300000], b[300000], c[300000], mn[300000], ans[300000];
int main() {
  int mni, mxi;
  scanf("%lld", &n);
  mxi = n - 1;
  mni = 0;
  memset(c, 0, sizeof(c));
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &b[i]);
    mn[i] = b[i];
    c[b[i]]++;
  }
  sort(mn, mn + n);
  for (int i = 0; i < n; i++) {
    if (c[n - a[i]]) {
      c[n - a[i]]--;
      ans[i] = 0;
    } else {
      while (!c[mn[mni]]) mni++;
      while (!c[mn[mxi]]) mxi--;
      int sz = lower_bound(mn, mn + mxi + 1, n - a[i]) - mn;
      while (!c[mn[sz]] && sz <= mxi) sz++;
      if (sz > mxi) {
        ans[i] = a[i] + mn[mni];
        c[mn[mni]]--;
      } else if (a[i] + mn[mni] < (a[i] + mn[sz]) % n) {
        ans[i] = a[i] + mn[mni];
        c[mn[mni]]--;
      } else {
        ans[i] = (a[i] + mn[sz]) % n;
        c[mn[sz]]--;
      }
    }
  }
  for (int i = 0; i < n; i++) printf("%d ", ans[i]);
  return 0;
}
