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:
//package rishab;
import java.io.*;
import java.util.*;
import java.util.Map.Entry;

public class Main {	
		public static void main(String[] args){  
			Scanner sc =new Scanner(System.in);
			int n=sc.nextInt();		
		   while(n-->0) {
			   int a=sc.nextInt();
			   String ans="";
			   if(a<=9) {
				   System.out.println(a);
			   }
			   else {  
				   for(int i=9;i>=1;i--) {
					   if(i<=a) {
						   a-=i;
						   ans=i+ans;
					   }
				   }
				   System.out.println((a>0)?-1:ans);
			   }
		   }
		}
		
		 static int octaltodecimal(int deciNum)
		    {
		        int octalNum = 0, countval = 1;
		        int dNo = deciNum;
		        while (deciNum != 0) {
		 
		            // decimals remainder is calculated
		            int remainder = deciNum % 8;
		 
		            // storing the octalvalue
		            octalNum += remainder * countval;
		 
		            // storing exponential value
		            countval = countval * 10;
		            deciNum /= 8;
		        }
		       return octalNum;
		    }
		
	  public static int floorSqrt(int x) 
	    { 
	        // Base Cases 
	        if (x == 0 || x == 1) 
	            return x; 
	  
	        // Do Binary Search for floor(sqrt(x)) 
	        int start = 1, end = x, ans=0; 
	        while (start <= end) 
	        { 
	            int mid = (start + end) / 2; 
	  
	            // If x is a perfect square 
	            if (mid*mid == x) 
	                return mid; 
	  
	            // Since we need floor, we update answer when mid*mid is 
	            // smaller than x, and move closer to sqrt(x) 
	            if (mid*mid < x) 
	            { 
	                start = mid + 1; 
	                ans = mid; 
	            } 
	            else   // If mid*mid is greater than x 
	                end = mid-1; 
	        } 
	        return ans; 
	    } 
		static int div(int n,int b) {
			int g=-1;
			for(int i=2;i<=Math.sqrt(n);i++) {
				if(n%i==0&&i!=b) {
					g=i;
					break;
				}
			}
			return g;
		}
	 
	  
	  public static int gcd(int a, int b) 
	    { 
	        if (a == 0) 
	            return b; 
	          
	        return gcd(b%a, a); 
	    } 
	   public static int lcm(int a, int b)  
	  {  
	     return (a*b)/gcd(a, b);  
	  }  
	   public static boolean isPrime1(int n) {
           if (n <= 1) {
               return false;
           }
           if (n == 2) {
               return true;
           }
           for (int i = 2; i <= Math.sqrt(n) + 1; i++) {
               if (n % i == 0) {
                   return false;
               }
           }
           return true;
       }
       public static boolean isPrime(int n) {
           if (n <= 1) {
               return false;
           }
           if (n == 2) {
               return true;
           }
           if (n % 2 == 0) {
               return false;
           }
           for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2) {
               if (n % i == 0) {
                   return false;
               }
           }
           return true;
       }
	      
	
}