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 MinimumArray {
    static TreeMap<Integer, Integer> b = new TreeMap<Integer, Integer>();
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int[] a = new int[n];
        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(st.nextToken());
        }
        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            add(Integer.parseInt(st.nextToken()));
        }
        for (int i = 0; i < n; i++) {
            int min = b.firstKey();
            int max = -1;
            if (n - a[i] <= b.lastKey()) max = b.ceilingKey(n - a[i]);
            if (max != -1) {
                System.out.print(Math.min((a[i] + min) % n, (a[i] + max) % n) + " ");
                if ((a[i] + max) % n <= (a[i] + min) % n) {
                    remove(max);
                } else {
                    remove(min);
                }
            } else {
                System.out.print((a[i] + min) % n + " ");
                remove(min);
            }
        }
    }

    static void add(int x){
        if(b.containsKey(x)){
            b.put(x, b.get(x) + 1);
        } else {
            b.put(x, 1);
        }
    }

    static void remove(int x){
        b.put(x, b.get(x) - 1);
        if(b.get(x) == 0){
            b.remove(x);
        }
    }
}
