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 <iostream>
#include <vector>
using namespace std;

int A[105];

int shift[105];

int N;

vector<int> SOLVE(void) {
	vector<int > IMP;
	IMP.push_back(-1);
	vector<int> V;
	for (int i = 1; i <= N; i++) {
		if (shift[i] < 0) { return IMP; }
	}

	for (int i = 1; i <= N; i++) {

		int K = -1;
		for (int j = 1; j <= N - i + 1; j++) {
			if (shift[j] == 0) { K = j; }
		}
		if (K == -1) { return IMP; }
		V.push_back(K);
		for (int h = K; h <= N - i; h++) {
			shift[h] = shift[h + 1] - 1;
		}

	}
	return V;
}

int main(void) {

	cin >> N;

	for (int i = 1; i <= N; i++) {
		cin >> A[i];
	}

	for (int i = 1; i <= N; i++) {
		shift[i] = i - A[i];
	}

	vector<int> W;
	W = SOLVE();

	int M = W.size();
	for (int i = 0; i < M; i++) {
		if (i != 0) { cout << " "; }
		cout << W[M-i-1];
	}
	cout << endl;

	//system("pause");
	return 0;
}