Mike has discovered a new way to encode permutations. If he has a permutation P = [p₁, p₂, ..., p_n], he will encode it in the following way:

Denote by A = [a₁, a₂, ..., a_n] a sequence of length n which will represent the code of the permutation. For each i from 1 to n sequentially, he will choose the smallest unmarked j (1 ≤ j ≤ n) such that p_i < p_j and will assign to a_i the number j (in other words he performs a_i = j) and will mark j. If there is no such j, he'll assign to a_i the number - 1 (he performs a_i = - 1).

Mike forgot his original permutation but he remembers its code. Your task is simple: find any permutation such that its code is the same as the code of Mike's original permutation.

You may assume that there will always be at least one valid permutation.

Input

The first line contains single integer n (1 ≤ n ≤ 500 000) — length of permutation.

The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ n or a_i = - 1) — the code of Mike's permutation.

You may assume that all positive values from A are different.

Output

In first and only line print n numbers p₁, p₂, ..., p_n (1 ≤ p_i ≤ n) — a permutation P which has the same code as the given one. Note that numbers in permutation are distinct.

Note

For the permutation from the first example:

i = 1, the smallest j is 2 because p₂ = 6 > p₁ = 2.

i = 2, there is no j because p₂ = 6 is the greatest element in the permutation.

i = 3, the smallest j is 1 because p₁ = 2 > p₃ = 1.

i = 4, the smallest j is 5 (2 was already marked) because p₅ = 5 > p₄ = 4.

i = 5, there is no j because 2 is already marked.

i = 6, the smallest j is 4 because p₄ = 4 > p₆ = 3.

Idx	Name	Rating
A	Mike and palindrome	1000
B	Mike and strings	1300
C	Mike and gcd problem	1700
D	Mike and distribution	2400
E	Mike and code of a permutation	3000

Codeforces Round 410 (Div. 2)