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)

/// Containers Start
#include <algorithm>
#include <bitset>
#include <cstdlib>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
/// C Header Files
#include <cassert>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <math.h>
#include <stdio.h>
/// Containers End

using namespace std;

/// Math Start
#define PI acos(-1.0)
#define Pi 3.141592653589793
#define EPS (1e-7)
#define INF (0x3f3f3f3f)
/// Math End

/// Extra Start
#define nn '\n'
#define pb push_back
#define ull unsigned long long
#define ll long long
#define MOD 1000000007
#define sz(a) int((a).size())
#define space " "
#define All(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
/// Extra End

#define rep(i, n) for (int i = 0; i < n; i++)
#define Cini(a)                                                                \
    int a;                                                                     \
    cin >> a;
#define Cinii(a, b)                                                            \
    int a, b;                                                                  \
    cin >> a >> b;
#define Ciniii(a, b, c)                                                        \
    int a, b, c;                                                               \
    cin >> a >> b >> c;
#define Cins(s)                                                                \
    string s;                                                                  \
    cin >> s;
#define Cinss(s1, s2)                                                          \
    string s1, s2;                                                             \
    cin >> s1 >> s2;
#define Cinc(c)                                                                \
    char c;                                                                    \
    cin >> c;
/// Functions End

/// Debug Start
#define deb(x) cout << #x << ": " << x << endl
#define deb2(x, y) cout << #x << ": " << x << '\t' << #y << ": " << y << endl;
#define deb3(x, y, z)                                                          \
    cout << #x << ": " << x << '\t' << #y << ": " << y << '\t' << #z << ": "   \
         << z << endl;
/// Debug End

int dx[] = {-1, 0, 1, 0};
int dy[] = {0, -1, 0, 1};

/**>>>>>>>>>>>>>>>>>>> END <<<<<<<<<<<<<<<<<<**/

/// template

signed main(void) {
    cin.sync_with_stdio(false);
    cin.tie(0);
    Cini(N);
    vector<int> b(N),c;
    unordered_map<int,int> m;
    rep(i,N) cin >> b[i];
    rep(i,N) m[b[i]]++;
    vector<int> res_vec(All(b));
    rep(i,N){
        vector<int> temp(All(res_vec));
        for(int j=N-1-i;j>=0;j--){
            if(j+1==temp[j]){
                c.push_back(temp[j]);
                temp.erase(temp.begin()+j);
                break;
            }
        }
        res_vec = temp;
    }
    if(res_vec.size()>0){
        cout << -1 << nn;
        return 0;
    }
    reverse(All(c));
    rep(i,N){
        cout << c[i] << nn;
    }
    return 0;
}