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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int main(){

    int N, i, j, ans=0, num, cnt=1, k;
    scanf("%d", &N);


    int a[N+1], b[N+1];
    for (i=1; i<=N; i++){
        scanf("%d", &b[i]);
    }


  int max, sequence[N+1], max_sequence;
  int *chk = (int*)calloc(N+1, sizeof(int));
 
  for (i=1; i<=N; i++){
    max = 0;
    for (j=N; j>=1; j--){
      if ((b[j] > max) && (chk[j] == 0)){
        max = b[j];
        max_sequence = j;
      }
    }
    sequence[i] = max_sequence;
    chk[max_sequence] = 1;
  } 

/*
    for (i=1; i<=N; i++){
        printf("%d (%d)\n", b[sequence[i]], sequence[i]);
    }

    return 0;*/

    int *picked = (int*)calloc(N+1, sizeof(int));
    int true_sequence[N+1], kcnt, N_;
  
    for (i=1; i<=N; i++){
        N_ = N-i+1;
        for (j=1; j<=N; j++){
            kcnt=0;
            for (k=1; k<=N; k++){//前から何番目か取得
                if (picked[k] == 0){
                    kcnt++;
                    true_sequence[k] = kcnt;
                }
            }
            //printf("%d (%d)\n", b[sequence[j]], sequence[j]);
            num = sequence[j];
            if (true_sequence[num] <= b[num] && picked[num] == 0 && b[num]<=N_){
                picked[num] = 1;
                a[cnt] = b[num];
                cnt++;
                //printf("%d (%d, %d)\n", b[num], num, true_sequence[num]);
                break;
            }
        }
    }
    

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

    return 0;
}