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 Go (1.6) (judge verdict: Wrong Answer)

package main

import "fmt"

// https://atcoder.jp/contests/agc032/tasks/agc032_a
func main() {
        var N int
        fmt.Scan(&N)

        b := []int{}
        for i := 0; i < N; i++ {
                var tmp int
                fmt.Scan(&tmp)
                b = append(b, tmp)
        }
        // sort
        bd := b
        for i := 0; i < N; i++ {
                for j := i; j < N; j++ {
                        if bd[i] < bd[j] {
                                tmp := bd[i]
                                bd[i] = bd[j]
                                bd[j] = tmp
                        }
                }
        }
        a := []int{}
        for i := 1; i <= N; i++ {
                l := len(bd)
                for j := 1; j <= l; j++ {
                        if bd[j-1] != -1 && i >= bd[j-1] {
                                a = append(a, bd[j-1])
                                bd[j-1] = -1
                                break
                        }
                }
        }
        if len(a) == len(b) {
                for i := 0; i < len(a); i++ {
                        fmt.Println(a[i])
                }
        } else {
                fmt.Print(-1)
        }
}