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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
const int N = 110;
int _w;

int n, a[N];
vector<int> ans;

int main() {
	cin >> n;
	for( int i = 1; i <= n; ++i )
		cin >> a[i];
	while(n) {
		bool flag = false;
		for( int i = n; i >= 1; --i )
			if( a[i] == i ) {
				ans.push_back(i);
				for( int j = i; j < n; ++j )
					a[j] = a[j+1];
				--n;
				flag = true;
				break;
			}
		if( flag == false ) {
			puts("-1");
			return 0;
		}
	}
	for( auto it = ans.rbegin(); it != ans.rend(); ++it )
		printf( "%d\n", *it );
	return 0;
}