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.util.*;

import static java.lang.Math.min;

public class Main {

    static int[] solve(int[] A, int[] B) {
        TreeMap<Integer, Integer> treeMap = new TreeMap<>();
        for (int i : B) {
            treeMap.put(i, treeMap.getOrDefault(i, 0) + 1);
        }
        int n = A.length;
        int[] C = new int[n];
        for (int i = 0; i < n; i++) {
            int k = n - A[i];

            Integer right = treeMap.higherKey(k - 1);
            Integer left = treeMap.firstKey();
            int remove = -1;
            if (right == null || (left + A[i]) % n <= (right + A[i]) % n) {
                C[i] = (left + A[i]) % n;
                remove = left;
            } else {
                C[i] = (right + A[i]) % n;
                remove = right;
            }
            if (treeMap.get(remove) == 1) {
                treeMap.remove(remove);
            } else {
                treeMap.put(remove, treeMap.get(remove) - 1);
            }


        }
        return C;

    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[] A = new int[n];
        int[] B = new int[n];
        for (int i = 0; i < n; i++) {
            A[i] = scanner.nextInt();
        }
        for (int i = 0; i < n; i++) {
            B[i] = scanner.nextInt();
        }
        int[] res = solve(A, B);
        StringBuilder s = new StringBuilder();
        for (int re : res) {
            s.append(re);
            s.append(" ");
        }
        System.out.println(s.toString());
    }
}



