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>

typedef unsigned char bool;
int main(){
  int n;
  scanf("%d",&n);
  int a[n+1],b[n+1],i;
  bool flag=1;
  for(i=1;i<=n;i++)
    {
      scanf("%d",&a[i]);
      if(a[i]>i)flag=0;
      b[i]=0;
    }
  if(flag==0)
    {
      printf("-1\n");
      return 0;
    }
  //  printf("1\n");
  int j,temp,count;
  for(i=n;i>0;i--)
    {
      count=0;
      for(j=i+1;j<=n;j++)
	{
	  if(a[i]>=a[j]) count++;
	}
      //      printf("%2d ",count);
      for(j=0;j<count;j++)
	{
	  if(b[a[i]+j]>a[i])
	    count++;
	  //	  printf("%d:%d:%d ",b[i+j],a[i],count);
	}
      j=a[i]+count;
      //      printf("%2d\n",j);
      while(b[j]!=0)j++;
      b[j]=a[i];
    }
  //  printf("\n",j);
  for(i=1;i<=n;i++)
    printf("%d\n",b[i]);
  return 0;
}