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: Time Limit Exceeded)

#include <stdio.h>

void In(int M, int N, int b[N], int n){
  int i;
  for(i = M - 1;i >= n;i--) b[i] = b[i-1];
  b[n-1] = n;
}

int Check(int N, int a[N], int b[N], int c[N], int l[N], int ans[N], int n){
  int i, j;
  int d[n-1];
  if(n == N + 1){
    for(i = 0;i < N;i++) if(a[i] != b[i]) return -1;
    return 1;
  }
  for(i = 0;i < n - 1;i++) d[i] = b[i];
  for(i = 1;i <= n;i++){
    if(c[i-1] < l[i-1]){
      c[i-1]++;
      In(n, N, b, i);
      ans[n-1] = i;
      if(Check(N, a, b, c, l, ans, n + 1) == 1) return 1;
      c[i-1]--;
      for(j = 0;j < n - 1;j++) b[j] = d[j];
    }
  }
  return -1;
}

void Do(int N){
  int i;
  int a[N], b[N], c[N], l[N], ans[N];
  for(i = 0;i < N;i++) l[i] = 0;
  for(i = 0;i < N;i++){
    scanf("%d", &a[i]);
    l[a[i]-1]++;
    c[i] = 0;
  }
  if(Check(N, a, b, c, l, ans, 1) == -1) printf("-1\n");
  else for(i = 0;i < N;i++) printf("%d\n", ans[i]);
}

int main(void){
  int N;
  scanf("%d", &N);
  Do(N);
  return 0;
}