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: Wrong Answer)

#include <bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define dbg(x) cerr << #x << " = " << x << endl
#define _ << ' ' <<
using namespace std;
using ll = long long;
using vi = vector<int>;
using ii = pair<int, int>;
using vii = vector<ii>;

int n, b[100];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 0; i < n; ++i)
        cin >> b[i];

    vi a, sol;
    int m = n - 1, k = -1;
    for (int i = 0; i < n; ++i)
    {
        int j = m;
        for (; j >= 0; --j)
        {
            if (b[j] <= k + 2)
            {
                a.insert(a.begin() + b[j] - 1, b[j]);
                sol.push_back(b[j]);
                k++;
                while (k >= 0 && a[k] == b[m]) m--, k--;
                break;
            }
        }
    }
    if (sol.size() != n)
        return cout << -1, 0;
    for (int x : sol)
        cout << x << '\n';
}