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(void)
{
  int N,i = 0,ans = 0,buf0 = 0,buf1 = 0,j = 0;
  scanf("%d",&N);
  int a[N],b[N];
  for(i = 0;i < N;i++) {
    scanf("%d",&b[i]);
  }

  while(j < N) {
    //printf("ans %d buf0 %d buf1 %d\n",ans,buf0,buf1);
    i = N - 1;
    ans = -1;
    buf0 = buf1 = 0;

    for(i = N - 1;i >= 0;i--) {
      //printf("b[%d] %d\n",i,b[i]);
      buf1 = b[i];

      if(b[i] == i + 1) {
        ans = b[i];
        a[j] = ans;
        b[i] = buf0;
        break;
      }
        //printf("a[%d] %d\n",i,a[i]);

      b[i] = buf0;
      buf0 = buf1;
    }

    if(ans == -1) break;

    j++;
  }

  if(ans == -1) {
    printf("-1\n");
  } else {
    i = N - 1;
    for(i = N - 1;i >= 0;i--) {
      printf("%d\n",a[i]);
    }
  }

  return 0;
}