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 
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# input = io.StringIO(os.read(0, os.fstat(0).st_size).decode()).readline
ii = lambda:int(input())
kk=lambda:map(int,input().split())
# k2=lambda:map(lambda x:int(x)-1, input().split())
ll=lambda:list(kk())
n = ii()

parents, rank = [-1]*n, [0]*n
loc = [i for i in range(n)]
def findParent(x):
	stack = []
	curr = x
	while parents[curr] != -1:
		stack.append(curr)
		curr = parents[curr]
	for v in stack:
		parents[v] = curr
	return curr
def union(x, y):
	best = None
	xP, yP = findParent(x), findParent(y)
	if rank[x] < rank[y]: best=parents[xP] = yP;
	elif rank[x] > rank[y]: best=parents[yP] = xP
	else:
		best=parents[yP] = xP
		rank[xP]+=1
	if values[loc[best]] == 0:
		loc[best] = loc[xP] if yP is best else loc[yP]

a = kk()
values = [0]*n
tbp=[]
for x in kk(): values[x]+=1
for i in range(n):
	if values[i] == 0: union(i, (i+1)%n)
for v in a:
	p = loc[findParent((n-v)%n)]

	tbp.append((v+p)%n)
	values[p]-=1
	if values[p] == 0: union(p, (p+1)%n)
print(*tbp)