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 TestClass {
    public static void main(String args[] ) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int a[] = new int[n];
        int b[] = new int[n];
        int i;
        st = new StringTokenizer(br.readLine());
        for(i=0;i<n;i++)
        {
            a[i] = Integer.parseInt(st.nextToken());
        }
        st = new StringTokenizer(br.readLine());
        for(i=0;i<n;i++)
        {
            b[i] = Integer.parseInt(st.nextToken());
        }
        br.close();
        int c[] = solve(n,a,b);
        for(i=0;i<n;i++)
            out.print(c[i] + " ");
        out.println();
        out.close();
    }
    
    static int[] solve(int n,int a[],int b[])
    {
        int i;
        TreeMap<Integer,Integer> tm = new TreeMap<>();
        int c[] = new int[n];
        
        for(i=0;i<n;i++)
        {
            tm.put(b[i],tm.getOrDefault(b[i],0)+1);
        }

        
        int k;
        for(i=0;i<n;i++)
        {
            Object o = tm.ceilingKey(n-a[i]);
            if(o == null)
                k = tm.firstKey();
            else
                k = (int) o;
            c[i] = (a[i] + k)%n;
            
            int val = tm.get(k);
            if(val == 1)
                tm.remove(k);
            else
                tm.put(k,val-1);
        }
        
        return c;
    }
}