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: Time Limit Exceeded)

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    val b = readLine()!!.split(" ").map(Integer::parseInt)
    val result = DP(n).del(0, b)
    if (result == null) {
        println(-1)
    } else {
        println(result.joinToString(separator = "\n"))
    }
}


class DP(val n: Int) {
    fun del(c: Int, b: List<Int>): List<Int>? {
        //println(b)
        if (b.isEmpty()) return listOf()
        if (c == n) return null
        val dd = mutableListOf<List<Int>>()
        for (i in b.indices) {
            if (b[i] == i + 1) {
                dd.add((del(c + 1, b.filterIndexed { index, value ->
                    index != i
                }) ?: continue) + (i + 1))
            }
        }
        if (dd.isEmpty()) {
            return null
        }
        return dd[0]
    }
}