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: Wrong Answer)

#include<stdio.h>

int main(void){
  
  int N,i,j,tmp;
  char a[100],b[100];
  int AbleToFinish = 0;
  
  scanf("%d",&N);
  for(i = 1;i <= N;i++){
    scanf("%d ",&b[i]);
  }
  
  for(i = 1;i <= N;i++){
    for(j = i + 1;j <= N;j++){
      if(b[i] > b[j]){
        tmp = b[i];
        b[i] = b[j];
        b[j] = tmp;
      }
    }
  }
  
  for(i = 1;i <= N;i++){
    if(b[i] <= i){
      a[i] = b[i];
    }else if(b[i] > i){
      AbleToFinish = 1;
      goto result;
    }
  }
  
  result:
  
  if(AbleToFinish == 0){
    for(i = 1;i <= N;i++){
      printf("%d ",a[i]);
    }
  }else{
    printf("-1");
  }
  return 0;
}