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>
#pragma comment(linker, "/STACK:10000000000")
using namespace std;
double f(long long a, long long b, long long c) {
  double res = (double)((b * 1.0 + c * 1.0) / (a * 1.0 + c * 1.0));
  return res;
}
signed main() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  multiset<long long> b;
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long vr;
  for (long long i = 0; i < n; i++) {
    cin >> vr;
    b.insert(vr);
  }
  vector<long long> ans(n, 0);
  for (long long i = 0; i < n; i++) {
    long long need = (n - a[i]) % n;
    multiset<long long>::iterator it = b.lower_bound(need);
    if (it == b.end()) {
      it = b.begin();
    }
    long long pol = *it;
    b.erase(it);
    ans[i] = (a[i] + pol) % n;
  }
  for (long long i = 0; i < n; i++) cout << ans[i] << " ";
}
