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>

int main()
{
    int n,b[100],i,j,k = 0,ans[100];

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

    for(j = 0; j < n;k++)
    {
        for (i = n - 1; i >= 0; i--)
        {
            if(i + 1 == b[i])
            {
                ans[k] = i + 1;
                break;
            }
        }
        if(i == -1)break;
        for (; i < n - 1; i++)
        {
            b[i] = b[i + 1];
        }
        n--;
    }

    if(n == 0)
    {
        for (i = k - 1; i >= 0; i--)
        {
            printf("%d ",ans[i]);
        }
    }
    else
    {
        printf("-1");
    }
    
    
}