Mike and code of a permutation

对于一个排列 $P=[p_1,p_2,\dots,p_n]$ 定义它的编码 $A$ 为：对于每个 $i$，找到第一个未被标记过的位置 $j$ 满足 $p_j>p_i$，令 $a_i=j$，并打上标记；若没有合法的，则 $a_i=-1$。 现给出一个编码 $A$（除 $-1$ 外各不相同），请输出一个合法的 $P$。

Mike has discovered a new way to encode permutations. If he has a permutation $ P=[p_{1},p_{2},...,p_{n}] $ , he will encode it in the following way: Denote by $ A=[a_{1},a_{2},...,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}&lt;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.

The first line contains single integer $ n $ ( $ 1<=n<=500000 $ ) — length of permutation. The second line contains $ n $ space-separated integers $ a_{1},a_{2},...,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.

In first and only line print $ n $ numbers $ p_{1},p_{2},...,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.

6
2 -1 1 5 -1 4

2 6 1 4 5 3

8
2 -1 4 -1 6 -1 8 -1

1 8 2 7 3 6 4 5

For the permutation from the first example: $ i=1 $ , the smallest $ j $ is $ 2 $ because $ p_{2}=6&gt;p_{1}=2 $ . $ i=2 $ , there is no $ j $ because $ p_{2}=6 $ is the greatest element in the permutation. $ i=3 $ , the smallest $ j $ is $ 1 $ because $ p_{1}=2&gt;p_{3}=1 $ . $ i=4 $ , the smallest $ j $ is $ 5 $ ( $ 2 $ was already marked) because $ p_{5}=5&gt;p_{4}=4 $ . $ i=5 $ , there is no $ j $ because $ 2 $ is already marked. $ i=6 $ , the smallest $ j $ is $ 4 $ because $ p_{4}=4&gt;p_{6}=3 $ .