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 = input()
a = map(int, raw_input().strip().split())
b = map(int, raw_input().strip().split())

f = [0] * n
for num in b: f[num] += 1

ptr = [0] * n
for i in xrange(n):
    if f[i] != 0:
        ptr[i] = i
        j = (i - 1) % n
        while f[j] == 0:
            ptr[j] = ptr[(j + 1) % n]
            j -= 1
            if j < 0: j += n

nb = []
for i in xrange(n):
    x = a[i]
    y = n - x
    if y >= n: y -= n

    if f[y] == 0:
        idx = ptr[y]
        stk = []
        while f[idx] == 0:
            stk.append(idx)
            idx = ptr[idx]

        if stk:
            ch, par = stk[-1], idx
            while stk:
                ch = stk[-1]
                ptr[ch] = ptr[par]
                par = ch
                stk.pop()

        nb.append(idx)
        f[idx] -= 1
        if f[idx] == 0:
            ptr[idx] = ptr[(idx + 1) % n]
    else:
        nb.append(y)
        f[y] -= 1
        if f[y] == 0:
            ptr[y] = ptr[(y + 1) % n]

c = [str((a[i] + nb[i]) % n) for i in xrange(n)]
print ' '.join(c)