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: Accepted)

#include <string>
#include <vector>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <map>
#include <list>
#include <set>
#include <numeric>
#include <queue>
#include <stack>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <climits>
#include <cfloat>
#include <ctime>
#include <complex>
#include <cassert>
#include <array>
#include <bitset> 
#include <unordered_map>

using namespace std;
typedef long long LL;
typedef pair<int,int> P;

int main() {
	int N;
	cin >> N;
	vector<int> b;
	vector<int> ret;
	for(int i=0;i<N;i++){
		int v;
		cin >> v;
		b.push_back(v);
	}
	for(int i=N;i>0;i--){
		bool ok=false;
		for(int j=i-1;j>=0;j--){
			if(b[j]==j+1)
			{
				//cerr << j+1 << endl;
				ret.push_back(j+1);
				b.erase(b.begin()+j);
				ok=true;
				break;
			}
		}
		if(!ok){
			cout << -1 << endl;
			return 0;
		}
	}
	//cerr << ret.size() << endl;
	reverse(ret.begin(),ret.end());
	for(auto v:ret)
	{
		cout << v << endl;
	}


	return 0;
}