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 
// Don't place your source in a package

import java.math.BigInteger;
import java.util.*;
import java.lang.*;
import java.io.*;



// Please name your class Main
public class Main {
    static FastScanner fs=new FastScanner();
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        int Int() {
            return Integer.parseInt(next());
        }

        long Long() {
            return Long.parseLong(next());
        }

        String Str(){
            return next();
        }
    }


    public static void main (String[] args) throws java.lang.Exception {
        PrintWriter out = new PrintWriter(System.out);


        int T=1;
        for(int t=0;t<T;t++){
            int n=Int();
            int A[]=new int[n];
            int B[]=new int[n];
            for(int i=0;i<n;i++){
                A[i]=Int()%n;
            }
            for(int i=0;i<n;i++){
                B[i]=Int()%n;
            }
            Solution sol=new Solution(out);
            sol.solution(A,B);
        }
        out.flush();

    }


    public static int Int(){
        return fs.Int();
    }
    public static long Long(){
        return fs.Long();
    }
    public static String Str(){
        return fs.Str();
    }

}






class Solution{
    PrintWriter out;
    public Solution(PrintWriter out){
        this.out=out;
    }



    public void solution(int A[],int B[]){
       int n=A.length;
       int res[]=new int[n];
       TreeMap<Integer,Integer>tree=new TreeMap<>();
       for(int i=0;i<B.length;i++){
           if(!tree.containsKey(B[i]))tree.put(B[i],1);
           else tree.put(B[i],tree.get(B[i])+1);
       }

       for(int i=0;i<A.length;i++){
           int first=tree.firstKey();

           int need=n-A[i];
           Integer ceil=tree.ceilingKey(need);

           if(ceil!=null){
               res[i]=(A[i]+ceil)%n;
               remove(tree,ceil);
           }
           else{
               res[i]=(A[i]+first)%n;
               remove(tree,first);
           }
       }



       for(int i:res){
           out.print(i+" ");
       }
    }

    public void remove(TreeMap<Integer,Integer>tree,int key){
        tree.put(key,tree.get(key)-1);
        if(tree.get(key)==0)tree.remove(key);
    }





}



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                      free bug dog
*/





