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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>

int pop(int n, int i, int a[])
{
    int val;
    int ptr;

    val = a[i];

    for(ptr = i; ptr < n - 1; ptr++) {
        a[ptr] = a[ptr + 1];
    }

    return val;
}

int solve(int n, int a[], int b[])
{
    int i, j;

    for (i = n - 1; i >= 0; i--) {
        a[i] = -1;
        for (j = i; j >= 0; j--) {
            // とりあえず大きい値があったらreject
            if (b[j] > i + 1) {
                return -1;
            }
            // あったら取り除く
            if (b[j] == j + 1) {
                a[i] = j + 1;
                pop(n, j, b);
                break;
            }
        }
        if (a[i] == -1) {
            return -1;
        }
    }

    return 0;
}

int main(int argc, char *argv[])
{
    int n;
    int a[100];
    int b[100];
    int i, j;
    int ret;

    scanf("%d", &n);

    for (i = 0; i < n; i++) {
        scanf("%d", &b[i]);
    }

    ret = solve(n, a, b);

    if (ret != 0) {
        printf("-1\n");
        return 0;
    }

    for (i = 0; i < n; i++) {
        printf("%d\n", a[i]);
    }

    return 0;
}