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 M = 1e9 + 7;
int main() {
  long long int n;
  cin >> n;
  long long int a[n + 2], i;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int b[n + 2];
  map<long long int, long long int> m;
  for (i = 0; i < n; i++) {
    cin >> b[i];
    m[b[i]]++;
  }
  vector<long long int> v;
  long long int j;
  for (i = 0; i < n; i++) {
    long long int x = n - a[i];
    auto it = m.lower_bound(x);
    if (it != m.end()) cout << (a[i] + (*it).first) % n << " ";
    (*it).second--;
    if ((*it).second == 0) {
      m.erase(it);
    }
    if (it == m.end()) {
      auto k = m.lower_bound(0);
      cout << (a[i] + (*k).first) % n << " ";
      (*k).second--;
      if ((*k).second == 0) {
        m.erase(k);
      }
    }
  }
  cout << endl;
}
