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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.TreeMap;
import java.util.StringTokenizer;
import java.util.Map;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Washoum
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        inputClass in = new inputClass(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        EMinimumArray solver = new EMinimumArray();
        solver.solve(1, in, out);
        out.close();
    }

    static class EMinimumArray {
        static Integer[] a;
        static Integer[] b;

        public void solve(int testNumber, inputClass sc, PrintWriter out) {
            int n = sc.nextInt();
            a = new Integer[n];
            b = new Integer[n];

            TreeMap<Integer, Integer> map = new TreeMap<>();
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            int x;
            for (int i = 0; i < n; i++) {
                x = sc.nextInt();
                map.putIfAbsent(x, 0);
                map.put(x, map.get(x) + 1);
            }
            int[] c = new int[n];
            int act;
            for (int i = 0; i < n; i++) {
                if (map.ceilingKey(n - a[i]) == null) {
                    act = map.firstKey();
                } else {
                    act = map.ceilingKey(n - a[i]);
                }
                if (map.get(act) > 1) {
                    map.put(act, map.get(act) - 1);
                } else {
                    map.remove(act);
                }
                c[i] = (a[i] + act) % n;
            }
            for (int i = 0; i < n; i++) {
                out.print(c[i] + " ");
            }
        }

    }

    static class inputClass {
        BufferedReader br;
        StringTokenizer st;

        public inputClass(InputStream in) {

            br = new BufferedReader(new InputStreamReader(in));
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

