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 i,j,k,N,l;
    scanf("%d",&N);
    int b[101]={0};
    for(i=0;i<N;i++){
        scanf("%d",&b[i]);
    }
    int ans[101]={0};

    int flag=0;
    int bflag=0;
    int out=0;

    for(j=0;j<N&&out==0;j++){
        for(i=0;i<N-j;i++){
            if(b[i]==i+1&&b[i]>ans[j]){
                ans[j]=b[i];
                k=i;
                flag+=1;
            }
        }
        
        if(bflag==flag){
            out=1;
            flag=0;
        }
        
        bflag=flag;
        /*
        printf("j=%d\n",j);
        for(l=0;l<N;l++){
            printf("%d ",b[l]);
        }
        printf("\n");
        */
        for(;k<N;k++){
            b[k]=b[k+1];
        }
        

    }

    if(flag==0){
        printf("-1\n");
    }

    if(flag!=0){
        for(k=0;k<N;k++){
            printf("%d\n",ans[N-k-1]);
        }
    }
    

    return 0;
}