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 
// This template code suggested by KT BYTE Computer Science Academy
//   for use in reading and writing files for USACO problems.

// https://content.ktbyte.com/problem.java

import java.util.*;
import java.io.*;

public class MinimumArray {

    static StreamTokenizer in;

    static int nextInt() throws IOException {
        in.nextToken();
        return (int) in.nval;
    }

    public static void main(String[] args) throws Exception {
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

        int n = nextInt();
        TreeMap<Integer, Integer> numValues = new TreeMap<>();

        int[] a = new int[n];
        int[] b = new int[n];

        for (int i = 0; i < n; i++) {
            a[i] = nextInt() % n;
            if (a[i] == 0) a[i] = n;
        }

        for (int i = 0; i < n; i++) {
            b[i] = nextInt() % n;
            if (numValues.containsKey(b[i])) numValues.put(b[i], numValues.get(b[i]) + 1);
            else numValues.put(b[i], 1);
        }

        for (int i = 0; i < n; i++) {
            int to_add = n - a[i];

            if (numValues.ceilingKey(to_add) != null) {
                int key_ciel = numValues.ceilingKey(to_add);

                out.print((a[i] + key_ciel) % n);
                int num = numValues.get(key_ciel) - 1;
                if (num == 0) numValues.remove(key_ciel);
                else numValues.put(key_ciel, num);
            }

            else {
                int key_floor = numValues.ceilingKey(0);

                out.print((a[i] + key_floor) % n);
                int num = numValues.get(key_floor) - 1;
                if (num == 0) numValues.remove(key_floor);
                else numValues.put(key_floor, num);
            }

            if (i != n-1) {
                out.print(" ");
            }
        }

        out.close();
    }
}


