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 o_o = 2e5 + 5;
const int mod = 1e9 + 7;
const int oo = 0x7fffffff;
const int sup = 0x80000000;
int a[o_o];
int b[o_o] = {0};
int c[o_o];
int main() {
  int n;
  cin >> n;
  std::set<int> d;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i] = n - a[i];
  }
  for (int i = 1; i <= n; i++) {
    int u;
    scanf("%d", &u);
    d.insert(u);
    b[u]++;
  }
  for (int i = 1; i <= n; i++) {
    set<int>::iterator x = d.end();
    x--;
    if (a[i] > (*x)) {
      c[i] = (n - a[i] + *d.begin()) % n;
      if (--b[*d.begin()] == 0) d.erase(d.begin());
      continue;
    }
    set<int>::iterator it = d.lower_bound(a[i]);
    c[i] = (*it - a[i]) % n;
    if (--b[*it] == 0) d.erase(it);
  }
  for (int i = 1; i <= n; i++) printf("%d ", c[i]);
}
