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.*;
public class Cony {

	public static void main(String[] args) throws IOException
	{
		BufferedReader jk = new BufferedReader(new InputStreamReader(System.in)) ;
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out)) ;
		StringTokenizer ana =new StringTokenizer(jk.readLine()) ;
		int n =Integer.parseInt(ana.nextToken()) ; 
		ana =new StringTokenizer(jk.readLine()) ;
		int a[]= new int[n] ; 
		int b[]= new int[n] ;
		TreeSet<Integer> tree = new TreeSet<>(); 
		int occ[]= new int[n] ;
		for(int i=0 ;i< n ;i++)
		{
			a[i]= Integer.parseInt(ana.nextToken()) ;
		}
		ana =new StringTokenizer(jk.readLine()) ;
		for(int i=0 ;i< n ;i++)
		{
			b[i]= Integer.parseInt(ana.nextToken()) ;
			occ[b[i]]++; 
			tree.add(b[i]) ;
		}
		for(int i=0 ;i<n ;i++)
		{
			int ele = a[i] ;
			int y = n-ele ; 
			if(tree.ceiling(y)!=null)
			{
				int e = tree.ceiling(y) ;
				occ[e]-- ; 
				if(occ[e]==0) tree.remove(e) ;
				a[i]+=(e) ; 
				a[i]%=n ;
			}
			else
			{
				int e = tree.ceiling(0) ;
				occ[e]-- ; 
				if(occ[e]==0) tree.remove(e) ;
				a[i]+=(e) ; 
				a[i]%=n ;
			}
		}
		for(int i=0 ;i< n ;i++)
		{
			out.print(a[i]+" ");
		}
		


		out.flush();

	}
}
