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>
#include<string.h>
#include<math.h>

typedef long long ll;

int main(){
  int n;
  int op[105];
  int i,j,s;
  int b[105];
  int cnt = 1;
  int flag = 0;

  scanf("%d", &n);
  for(i = 1;i <= 104;i++) b[i] = -1;
  for(i = 1;i <= n;i++){
    scanf("%d", &b[i]);
  }

  for(i = 1;i <= n;i++){
    cnt = 1;
    for(j = 1;j <= n;j++){
      if(cnt == b[j]){
        s = j;
      }

      if(b[j] != -1) cnt++;
    }

    op[i] = b[s];
    b[s] = -1;
  }

  for(i = 1;i <= n;i++) if(b[i] != -1) flag = 1;

  if(flag != 1){
    for(i = n;i >= 1;i--) printf("%d\n", op[i]);
  }

  else printf("-1\n");

  return 0;
}