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,i;
    scanf("%d",&n);
    int b[101],j;
    b[n]=0;
    for(i=0;i<n;i++){
        scanf("%d",&b[i]);
    }
    
    int l=n;
    int op[100],k=0,flag=1;
    while(flag>0&&n>=0){
        flag=0;
        for(i=n-1;i>=0;i--){
            if(b[i]==i+1){
                op[k]=b[i];
                k++;
                flag=1;
                for(j=i;j<n-1;j++){
                    b[j]=b[j+1];
                }
                n--;
                break;
            }
        }
    }
    if(k<l){
        printf("-1");
        return 0;
    }
    for(i=k-1;i>=0;i--){
    printf("%d\n",op[i]);
    }
    return 0;
}