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: Time Limit Exceeded)

#include <stdio.h>


void stack(int *b, int *a, int m){
    int i, j, flag;
    int new_b[100];

    if(m==1){
        if(b[0] == 1){
            a[0] = 1;
        }else{
            a[0] = -1;
        }
    }else{
        flag = 0;
        for(i=0;i<m;i++){
            if(b[i] == i+1){
                // for(j=0;j<i;j++){
                //    new_b[j] = b[j];
                // }
                for(j=i+1;j<m;j++){
                    new_b[j-1] = b[j];
                }
                stack(new_b, a, m-1);
                if(a[m-2] == -1){
                    new_b[i] = b[i];
                    continue;
                }else{
                    a[m-1] = i+1;
                    flag = 1;
                }
            }
            new_b[i] = b[i];
        }
        if(flag == 0){
            a[m-1] = -1;
        }
    }
}

int main(void)
{
    int n;
    int b[100];

    int i;
    int a[100];

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

    stack(b, a, n);

    if(a[n-1] == -1){
        printf("%d", -1);
    }else{
        for(i=0;i<n;i++){
            printf("%d\n", a[i]);
        }
    }

    return 0;
}