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