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: Accepted)

const val MOD: Long = (1e9+7).toLong()

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    /* val b = ArrayList(readLine()!!.split(' ').map{it.toInt()-1}) */
    val b = arrayListOf(*readLine()!!.split(' ').map{it.toInt()-1}.toTypedArray())

    val l = arrayListOf(0)
    l.removeAt(0)
    /* val l: ArrayList<Int> = arrayListOf() */
    while (b.size > 1) {
        var f = false
        for ((i, x) in b.withIndex().reversed()) {
            if (i == 0) continue
            if (i == x) {
                l.add(x+1)
                b.removeAt(i)
                f = true
                break
            }
        }

        if (!f) {
            if (b[0] == 0) {
                l.add(1)
                b.removeAt(0)
                continue
            }

            println(-1)
            return
        }
    }
    if (b[0] != 0) {
        println(-1)
        return
    } else {
        println(1)
        for (x in l.reversed()) {
            println(x)
        }
    }
}