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>
using namespace std;

int main(){
        int n;
        scanf("%d", &n);
        list<int> a;
        list<int> ::iterator it;
        int num;
        for(int i=0; i<n; i++) { scanf("%d", &num);a.push_back(num); }

        stack<int> s;
        for(int i=0; i<n; i++){
                int j=1;
                for(it=a.begin(); it!=a.end(); it++, j++){
                        if(*it == j) continue;
                        else if( j == 1 ) { printf("-1\n"); return 0; }
                        else { it--; s.push(*it); it = a.erase(it); break; }
                }
                if(j == n-i) break;
        }

        for(it=a.begin(); it!=a.end(); it++) printf("%d\n", *it);
        while(!s.empty()){ printf("%d\n", s.top()); s.pop();};

        return 0;
}