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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
	static int[] ans;
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		String b = br.readLine().replaceAll(" ", "");

		ans = new int[n];
		boolean flg = dfs("", n, b, ans);
		if (flg) {
			for (int i = 0; i < n; i++) {
				System.out.println(ans[i]);
			}
		} else {
			System.out.println(-1);
		}
	}

	static boolean dfs(String str, int n, String b, int[] ans) {
		if (str.length() == n) {
			if (str.equals(b)) {
				return true;
			}
			return false;
		}
		for (int i = 1; i <= str.length() + 1; i++) {
			ans[str.length()] = i;
			boolean flg = dfs(str.substring(0, i - 1) + i + str.substring(i - 1, str.length()), n, b, ans);
			if (flg) {
				return true;
			}
		}
		return false;
	}
}