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 9.2.1) (judge verdict: Accepted)

#pragma region template 2.4
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using pq_asc = priority_queue<T, vector<T>, greater<T>>;
typedef long long ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef pair<ll, ll> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
#define REP(i, n) for (ll i = 0; i < (n); ++i)
#define REP1(i, n) for (ll i = 1; i <= (n); ++i)
#define FOR(i, a) for (auto &i : a)
#define CH(f, x, y) x = f(x, y)
#define IN(T, x) \
    T x;         \
    cin >> x;
#define AIN(T, a, n) \
    vector<T> a(n);  \
    FOR(i, a)        \
    cin >> i;
#define A2IN(T1, a, T2, b, n) \
    vector<T1> a(n);          \
    vector<T2> b(n);          \
    REP(i, n)                 \
    cin >> a[i] >> b[i];
#define OUT(x) cout << (x) << endl;
#define FOUT(x) cout << fixed << setprecision(15) << (x) << endl;
#define ALL(a) (a).begin(), (a).end()
#define SORT(a) sort(ALL(a))
#define RSORT(a) \
    SORT(a);     \
    reverse(ALL(a))
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define DUMPA(a)          \
    cout << #a << " = {"; \
    JOUT(ALL(a), ", ", cout) << "}" << endl;

template <typename T>
ostream &JOUT(T s, T e, string sep = " ", ostream &os = cout)
{
    if (s != e)
    {
        os << *s;
        ++s;
    }
    while (s != e)
    {
        os << sep << *s;
        ++s;
    }
    return os;
}

ostream &YES(bool cond, string yes = "Yes", string no = "No", ostream &os = cout)
{
    if (cond)
    {
        os << yes << endl;
    }
    else
    {
        os << no << endl;
    }
    return os;
}

template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p)
{
    os << '(' << p.first << ", " << p.second << ')';
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v)
{
    os << '[';
    JOUT(ALL(v), ", ", os) << ']';
    return os;
}

const ll INF = 1e18;
const ll MOD = 1e9 + 7;
#pragma endregion template

int main()
{
    IN(ll, N);
    AIN(ll, b, N);
    vi ans;
    for (ll i = N; i > 0; --i)
    {
        bool ok = false;
        for (ll j = i; j > 0; --j)
        {
            if (b[j - 1] == j)
            {
                ans.push_back(j);
                b.erase(b.begin() + (j - 1));
                ok = true;
                break;
            }
        }
        if (not ok)
        {
            OUT(-1);
            return 0;
        }
    }
    for (ll i = N - 1; i >= 0; --i)
    {
        OUT(ans[i]);
    }
}