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)

#include<cstdio>
#include<vector>
#include<string>
#include<cmath>

using namespace std;

int main()
{
	int n;
	vector<int> arr;
	vector<int> sol;
	int input;
	int i;
	bool valid;

	scanf("%d", &n);

	for (i = 0; i < n; i++)
	{
		scanf("%d", &input);

		arr.push_back(input);
	}

	while (!arr.empty())
	{
		valid = false;

		for (i = arr.size()-1; i >= 0; i--)
		{
			if (arr[i] == i + 1)
			{
				sol.insert(sol.begin(), arr[i]);

				arr.erase(arr.begin() + i);

				valid = true;

				break;
			}
		}
		
		if (!valid)
		{
			printf("-1\n");

			return 0;
		}
	}

	for (i = 0; i < n; i++)
	{
		printf("%d\n", sol[i]);
	}

	return 0;
}