CF1054B.Appending Mex

Initially Ildar has an empty array. He performs $n$ steps. On each step he takes a subset of integers already added to the array and appends the mex of this subset to the array.

The mex of an multiset of integers is the smallest non-negative integer not presented in the multiset. For example, the mex of the multiset $[0, 2, 3]$ is $1$ , while the mex of the multiset $[1, 2, 1]$ is $0$ .

More formally, on the step $m$ , when Ildar already has an array $a_1, a_2, \ldots, a_{m-1}$ , he chooses some subset of indices $1 \leq i_1 < i_2 < \ldots < i_k < m$ (possibly, empty), where $0 \leq k < m$ , and appends the $mex(a_{i_1}, a_{i_2}, \ldots a_{i_k})$ to the end of the array.

After performing all the steps Ildar thinks that he might have made a mistake somewhere. He asks you to determine for a given array $a_1, a_2, \ldots, a_n$ the minimum step $t$ such that he has definitely made a mistake on at least one of the steps $1, 2, \ldots, t$ , or determine that he could have obtained this array without mistakes.

The first line contains a single integer $n$ ( $1 \leq n \leq 100\,000$ ) — the number of steps Ildar made.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ( $0 \leq a_i \leq 10^9$ ) — the array Ildar obtained.

If Ildar could have chosen the subsets on each step in such a way that the resulting array is $a_1, a_2, \ldots, a_n$ , print $-1$ .

Otherwise print a single integer $t$ — the smallest index of a step such that a mistake was made on at least one step among steps $1, 2, \ldots, t$ .

In the first example it is possible that Ildar made no mistakes. Here is the process he could have followed.

• $1$ -st step. The initial array is empty. He can choose an empty subset and obtain $0$ , because the mex of an empty set is $0$ . Appending this value to the end he gets the array $[0]$ .
• $2$ -nd step. The current array is $[0]$ . He can choose a subset $[0]$ and obtain an integer $1$ , because $mex(0) = 1$ . Appending this value to the end he gets the array $[0,1]$ .
• $3$ -rd step. The current array is $[0,1]$ . He can choose a subset $[0,1]$ and obtain an integer $2$ , because $mex(0,1) = 2$ . Appending this value to the end he gets the array $[0,1,2]$ .
• $4$ -th step. The current array is $[0,1,2]$ . He can choose a subset $[0]$ and obtain an integer $1$ , because $mex(0) = 1$ . Appending this value to the end he gets the array $[0,1,2,1]$ .

Thus, he can get the array without mistakes, so the answer is $-1$ .

In the second example he has definitely made a mistake on the very first step, because he could not have obtained anything different from $0$ .

In the third example he could have obtained $[0, 1, 2]$ without mistakes, but $239$ is definitely wrong.