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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
void prnt(vector<int> b){
	for(int i=0;i<b.size();i++){
		cout<<b[i]<<endl;
	}
	return;
}
void ok(vector<int>& b,vector<int>& ans,int* bo){
	//prnt(b);
	//cout<<"size="<<b.size()<<endl;
	if(b.size()==1&&b[0]==1){
		//cout<<"hoge\n";
		ans.push_back(1);
		*bo=1;
		return;
	}
	for(int i=b.size()-1;i>=0;i--){
		if(b[i]==i+1){
			//cout<<"ho\n";
			b.erase(b.begin()+i);
			ans.push_back(i+1);
			ok(b,ans,bo);
			if(*bo==1) return;
			ans.pop_back();
			b.insert(b.begin()+i,i+1);
		}
	}
	return ;
}

int main(){
	int n;
	cin>>n;
	vector<int> b(n,0);
	vector<int> ans;
	for(int i=0;i<n;i++){
		cin>>b[i];
	}
	int bo=0;
	ok(b,ans,&bo);
	//cout<<"bo="<<bo<<endl;
	if(bo==1){
		for(int i=0;i<ans.size();i++){
			cout<<ans[ans.size()-i-1]<<endl;
		}
	}else{
		cout<<-1<<endl;
	}
	return 0;
}