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 
n = int(input())
a = map(int,input().split())
b = map(int,input().split())
parent, rank = [-1] * n, [0] * n
source = [i for i in range(n)]
occurences = [0]*n
printer = []
def fp(x):
    stack = []
    curr = x
    while parent[curr] != -1:
        stack.append(curr)
        curr = parent[curr]
    for v in stack:
        parent[v] = curr
    return curr
def u(x, y):
    best = None
    xP, yP = fp(x), fp(y)
    if rank[x] < rank[y]: best=parent[xP] = yP;
    elif rank[x] > rank[y]: best=parent[yP] = xP
    else:
        best=parent[yP] = xP
        rank[xP]+=1
    if occurences[source[best]] == 0:
        source[best] = source[xP] if yP is best else source[yP]
for x in b: occurences[x]+=1
for i in range(n):
    if occurences[i] == 0: u(i, (i+1)%n)
for val in a:
    p = source[fp((n-val)%n)]
    printer.append(str((val+p)%n))
    occurences[p]-=1
    if occurences[p] == 0: u(p, (p+1)%n)
print(" ".join(printer))