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 mnarr {
    public static void main( String[] args ) {
	Scanner in = new Scanner( System.in );
	int n = in.nextInt();
	int[] a = new int[n];
	for ( int i = 0; i < n; i++ )
	    a[i] = in.nextInt();
	TreeMap<Integer, Integer> b = new TreeMap<>();
	for ( int i = 0; i < n; i++ ) {
	    int bi = in.nextInt();
	    if ( !b.containsKey( bi ) )
		b.put( bi, 0 );
	    b.put( bi, b.get( bi ) + 1 );
	}
	for ( int i = 0; i < n; i++ ) {
	    Integer c = b.higherKey( n - 1 - a[i] );
	    if ( c == null )
		c = b.firstKey();
	    System.out.print( ( ( a[i] + c ) % n ) + " " );
	    if ( b.get( c ) > 1 )
		b.put( c, b.get( c ) - 1 );
	    else
		b.remove( c );
	}
    }
}
