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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
int N;
bool fin = false;

int search(vector<int> xx, vector<int> ans){
    // int check = 1;
    // bool flag = false;
    if(xx.size() == 1){
        for(int i = ans.size()-1; i >= 0; i--){
            cout << ans.at(i) << endl;
        }
        fin = true;
        return 0;
    }
    for(int i = 1; i < xx.size() ; i++){
        if(fin){
            return 0;
        }
        if(xx.at(i) == i){
            //erase
            // flag = true;
            vector<int> new_xx = xx;
            new_xx.erase(new_xx.begin() + i);

            ans.push_back(i);
            search(new_xx, ans);
            // check *= search(new_xx, ans);
            ans.pop_back();
        }
    }
    return 0;
}

int main(){
    cin >> N;
    vector<int> b = vector<int>(N+1);
    for(int i = 1; i<N+1; i++){
        cin >> b.at(i);
    }
    vector<int> ans;
    search(b, ans);
    if(fin == false){
        cout << -1 << endl;
    }
    return 0;
}