Problem: You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).

Input

The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.

Each test case consists of a single integer number x (1 ≤ x ≤ 50).

Output

Output t answers to the test cases:

  * if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number; 
  * otherwise print -1. 

Example

Input


4
1
5
15
50


Output


1
5
69
-1

Solution (in JAVA):

Solution:


import java.util.Scanner;

public class UniqueNumber {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();

		while (t-- > 0) {
			int n = sc.nextInt();
			if (n > 45)
				System.out.println("-1");
			else {
				if (n >= 1 && n <= 9)
					System.out.println(n);
				else if (n >= 10 && n <= 17)
					System.out.println(((n - 9) * 10 + 9));
				else if (n >= 18 && n <= 24)
					System.out.println((n - 9 - 8) * 100 + 8 * 10 + 9);
				else if (n >= 25 && n <= 30)
					System.out.println((n - 9 - 8 - 7) * 1000 + 7 * 100 + 8 * 10 + 9);
				else if (n >= 31 && n <= 35)
					System.out.println((n - 9 - 8 - 7 - 6) * 10000 + 6 * 1000 + 7 * 100 + 8 * 10 + 9);
				else if (n >= 36 && n <= 39)
					System.out.println((n - 9 - 8 - 7 - 6 - 5) * 100000 + 5 * 10000 + 6 * 1000 + 7 * 100 + 8 * 10 + 9);
				else if (n >= 40 && n <= 42)
					System.out.println((n - 9 - 8 - 7 - 6 - 5 - 4) * 1000000 + 4 * 100000 + 5 * 10000 + 6 * 1000
							+ 7 * 100 + 8 * 10 + 9);
				else if (n >= 43 && n <= 44)
					System.out.println((n - 9 - 8 - 7 - 6 - 5 - 4 - 3) * 10000000 + 3 * 1000000 + 4 * 100000 + 5 * 10000
							+ 6 * 1000 + 7 * 100 + 8 * 10 + 9);
				else if (n == 45)
					System.out.println(123456789);

			}
		}
	}

}