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)

//#pragma GCC target ("avx2")
//#pragma GCC optimize ("Ofast")
//#pragma GCC optimize ("unroll-loops")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define int long long

using namespace std;
using namespace __gnu_pbds;

template <class T> using ordered_set =
tree<T, null_type, less<T>,
rb_tree_tag, tree_order_statistics_node_update>;

const int N = (int)(1e6 + 5);
const int M = (int)(1e9 + 7);
const int INF = (int)(1e18 + 9);
const int dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
const int dy[] = {0, 0, 1, -1, 1, -1, 1, -1};

int a[N];

signed main() {
  ios_base :: sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector <int> v;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    v.push_back(x);
  }
  vector <int> ret;
  while (!v.empty()) {
    bool ch = 0;
    for (int i = (int)v.size() - 1; i >= 0; i--) {
      if (v[i] == i + 1) {
        ch = 1;
        ret.push_back(v[i]);
        v.erase(v.begin() + i);
        break;
      }
    }
    if (!ch) {
      cout << "-1";
      return 0;
    }
  }
  reverse(ret.begin(), ret.end());
  for (auto x : ret)
    cout << x << "\n";
  return 0;
}