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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class CF1157E {
    static TreeMap<Integer,Integer> b;
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        int [] a = new int[n];
        b = new TreeMap<>();
        StringTokenizer s = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(s.nextToken());
        }
        s = new StringTokenizer(br.readLine());
        for (int i = 0; i < n; i++) {
            addToB(Integer.parseInt(s.nextToken()));
        }
        int [] res = new int[n];
        for (int i = 0; i < n; i++) {
            int ideal = 0;
            if (a[i] == 0) {
                ideal = 0;
            }
            else {
                ideal = n-a[i];
            }
            Integer match1 = b.ceilingKey(ideal);
            Integer match2 = b.ceilingKey(0);
            int bestMatch = 0;
            if (match1 != null) {
                if ((a[i]+match1)%n < (a[i]+match2)%n) {
                    bestMatch = match1;
                }
                else {
                    bestMatch = match2;
                }
            }
            else {
                bestMatch = match2;
            }
            res[i] = (bestMatch + a[i]) % n;
            removeFromB(bestMatch);
        }
        for (int num : res) System.out.print(num+" ");
    }
    static void addToB (int num) {
        if (!b.containsKey(num)) b.put(num,1);
        else b.put(num,b.get(num)+1);
    }
    static void removeFromB (int num) {
        b.put(num,b.get(num)-1);
        if (b.get(num)==0) b.remove(num);
    }
}
