Problem:
Score : 400 points

Problem Statement

Snuke has an empty sequence a.

He will perform N operations on this sequence.

In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).

You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq b_i \leq N

Input

Input is given from Standard Input in the following format:

N
b_1 \dots b_N

Output

If there is no sequence of N operations after which a would be equal to b, print -1.
If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.

Sample Input 1

3
1 2 1

Sample Output 1

1
1
2

In this sequence of operations, the sequence a changes as follows:

After the first operation: (1)

After the second operation: (1,1)

After the third operation: (1,2,1)

Sample Input 2

2
2 2

Sample Output 2

-1

2 cannot be inserted at the beginning of the sequence, so this is impossible.

Sample Input 3

9
1 1 1 2 2 1 2 3 2

Sample Output 3

1
2
2
3
1
2
2
1
1

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::{input, fastout};

#[fastout]
fn main() {
    input! {
        n: usize,
        bb: [usize; n],
    }

    for (i, &b) in bb.iter().enumerate() {
        if i + 1 < b {
            println!("-1");
            return;
        }
    }

    let mut ans = vec![];
    f(&mut ans, &bb, 1);
    for i in ans {
        println!("{}", i);
    }
}

fn f(ans: &mut Vec<usize>, seq: &[usize], n: usize) {
    let mut vec = vec![];
    for (i, &v) in seq.iter().enumerate() {
        if v == n {
            vec.push(i);
        }
    }
    let mut prev = None;
    for &i in vec.iter().rev() {
        ans.push(n);
        if let Some(prev) = prev {
            f(ans, &seq[i..prev], n+1);
        } else {
            f(ans, &seq[i..], n+1);
        }
        prev = Some(i);
    }
}