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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*
fun main(args: Array<String>){
	val cin = Scanner(System.`in`)
    val n = cin.nextInt()
    val b = Array<MutableList<Int>>(n, { mutableListOf() })

    for (i in 0..n-1) {
        val x = cin.nextInt() - 1
        b[x].add(i)
    }

    val ans = Array<Int>(n, { -1 })

    var all_ok = true

    for (i in 0..n-1) {
        val j = n - 1 - i
        for (e in b[j]) {
            var gone = e - j
            var ok = false

            var now = n - 1
            var now_gone = 0
            while (gone >= 0 && n > gone && now >= 0) {
                
                if (now_gone == gone && ans[now] == -1) {
                    ans[now] = j
                    ok = true
                    break
                }
                if (ans[now] == -1) {
                    now_gone += 1
                }
                now -= 1
            }
            if (!ok) {
                all_ok = false
                break
            }
        }

        if (!all_ok) {
            break
        }
    }

    if (!all_ok) {
        println(-1)
    } else {
        for (i in 0..n-1) {
            println(ans[i]+1)
        }
    }
}