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 

import java.io.*;
import java.util.Scanner;
import java.util.*;
public class MinimumArray {
	public static void main(String[] args){
		Scanner input = new Scanner(System.in);
		TreeMap<Integer,Integer> b = new TreeMap<>();
		int n = input.nextInt();
		int[] a = new int[n];
		int[] c = new int[n];
		for(int i = 0; i < n; i++){
			a[i] = input.nextInt();
		}
		for(int i = 0; i < n; i++){
			int temp = input.nextInt();
			if(b.containsKey(temp)){
				int temp2 = b.get(temp);
				b.put(temp,temp2 + 1);
			}
			else{
				b.put(temp,1);
			}
		}
		for(int i = 0; i < n; i++){
			int min = b.firstKey();
			int max = b.lastKey();
			if((n-a[i]) > max || (n-a[i]) < min){
				c[i] = (a[i] + min)%n;
				b.put(min,b.get(min) - 1);
				if(b.get(min) == 0){
					b.remove(min);
				}

			}
			else{
				SortedMap<Integer, Integer> temp = new TreeMap<Integer,Integer>();
				temp = b.tailMap(n-a[i]);
				int d = temp.firstKey();
				c[i] = (a[i] + d)%n;
				b.put(d,b.get(d) - 1);
				if(b.get(d) == 0) {b.remove(d);}
			}
			System.out.print(c[i] + " ");
		}

	}
}