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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main() {
  int n;
  cin >> n;

  vector<int> a(n);
  vector<int> b(n);

  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }

  for (int i = 0; i < n; i++) {
    int j;
    for (j = n - i - 1; j >= 0; j--) {
      if (b[j] == j + 1) {
        a[i] = j + 1;
        for (int k = j; k < n - i - 1; k++) {
          b[k] = b[k + 1];
        }
        break;
      }
    }
    if (j < 0) {
      cout << "-1" << endl;
      return 0;
    }
  }

  reverse(a.begin(), a.end());
  for (int x : a) {
    cout << x << endl;
  }

  return 0;
}