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.*;

public class F {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int a[] = new int[n];
		for(int i=0;i<n;i++)
			a[i] = sc.nextInt();
		TreeMap<Integer, Integer> map = new TreeMap<>();
		for(int i=0;i<n;i++) {
			int num = sc.nextInt();
			map.put(num, map.getOrDefault(num, 0) + 1);
		}
		
		StringBuilder sb = new StringBuilder();
		for(int i = 0;i<n;i++) {
			Integer num = map.ceilingKey(n - a[i]);
			if(num == null) {
				num = map.ceilingKey(0);		
			}
			
			int f = map.get(num);
			if(f == 1) {
				map.remove(num);
			}
			else {
				map.put(num, f-1);
			}
			sb.append((num+a[i]) % n + " ");
		}
		System.out.println(sb);
		
	}

}
