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 
/*
roses are red
memes are neat
all my test cases time out
lmao yeet
*/
import java.util.*;
import java.io.*;

   public class E
   {
      public static void main(String args[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         int[] arr1 = new int[N];
         st = new StringTokenizer(infile.readLine());
         for(int i=0; i < N; i++)
            arr1[i] = Integer.parseInt(st.nextToken());
         st = new StringTokenizer(infile.readLine());
         int[] arr2 = new int[N];
         for(int i=0; i < N; i++)
            arr2[i] = Integer.parseInt(st.nextToken());
         //If I get this problem I will steal george's girl
         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
         TreeSet<Integer> bst = new TreeSet<Integer>();
         for(int i=0; i < N; i++)
         {
            int val = arr2[i];
            if(!bst.contains(val))
            {
               bst.add(val);
               map.put(val, 0);
            }
            else
               map.put(val, map.get(val)+1);
         }
         int[] res = new int[N];
         for(int i=0; i < N; i++)
         {
            int check = N-arr1[i];
            int boof;
            if(bst.last() >= check)
               boof = bst.ceiling(check);
            else
            {
               boof = bst.first();
            }
            res[i] = (arr1[i]+boof)%N;
            bst.remove(boof);
            if(map.get(boof) > 0)
            {
               bst.add(boof);
               map.put(boof, map.get(boof)-1);
            }
         }
         StringBuilder sb = new StringBuilder();
         for(int i=0; i < N; i++)
         {
            if(i == N-1)
               sb.append(res[i]);
            else
               sb.append(res[i]+" ");
         }
         System.out.println(sb);
      }
   }