description: In Berland, n different types of banknotes are used. Banknotes of the i-th type have denomination 10^{a_i} burles (burles are the currency used in Berland); the denomination of banknotes of the first type is exactly 1.
  
  Let's denote f(s) as the minimum number of banknotes required to represent exactly s burles. For example, if the denominations of banknotes used in Berland are 1, 10 and 100, then f(59) = 14: 9 banknotes with denomination of 1 burle and 5 banknotes with denomination of 10 burles can be used to represent exactly 9 ⋅ 1 + 5 ⋅ 10 = 59 burles, and there's no way to do it with fewer banknotes.
  
  For a given integer k, find the minimum positive number of burles s that cannot be represented with k or fewer banknotes (that is, f(s) > k).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
  
  The first line of each test case contains two integers n and k (1 ≤ n ≤ 10; 1 ≤ k ≤ 10^9).
  
  The next line contains n integers a_1, a_2, ..., a_n (0 = a_1 < a_2 < ... < a_n ≤ 9).
  
  Output
  
  For each test case, print one integer — the minimum positive number of burles s that cannot be represented with k or fewer banknotes.
  
  Example
  
  Input
  
  
  4
  3 13
  0 1 2
  2 777
  0 4
  3 255
  0 1 3
  10 1000000000
  0 1 2 3 4 5 6 7 8 9
  
  
  Output
  
  
  59
  778
  148999
  999999920999999999
solution: import java.io.*;
  
  
  
  	import java.util.*;
  	
  
  	public class Aqueous {
  		
  		static MyScanner sc = new MyScanner();
  		
  		public static void main(String[] args) {
  			int t = sc.nextInt();
  			
  			while(t-->0) {
  				
  				int n = sc.nextInt();
  				long k = sc.nextLong();
  				int a[] = new int[n];
  				long ar[] = new long[n];
  				
  				for(int i= 0; i<n; i++) {
  					a[i] = sc.nextInt();
  					ar[i] = (long)Math.pow((double)10, (double)a[i]);
  				}
  				
  				k++;
  				
  				long num = 0;
  				
  				for(int i= 0; i<n-1; i++) {
  					
  					
  					if(k<=9) {
  						num += k*(ar[i]);
  						break;
  					}
  					long times = ar[i+1]/ar[i];
  					times--;
  					if(times>=k) {
  						num = num+(k)*ar[i];
  						break;
  					}
  					num = num + (times)*ar[i];
  					k-= times;
  					
  					if(i==n-2) {
  						num += ar[i+1]*k;
  						break;
  					}
  				}
  				
  				if(n==1) {
  					
  					long ans = k/ar[0];
  					if(k%ar[0]!=0) {
  						ans++;
  					}
  					System.out.println(ans);
  				}
  				else {
  				System.out.println(num);
  				}
  			}
  		}
  		
  		
  		static void ruffleSort(int a[]) {
  
  	          int n = a.length;
  
  	          Random r = new Random();
  
  	          for (int i = 0; i < n; i++) {
  
  	              int oi = r.nextInt(n);
  	              int temp = a[oi];
  	              a[oi] = a[i];
  	              a[i] = temp;
  	          }
  
  	          Arrays.sort(a);
  	      }
  				
  		static void ruffleSort(long a[]) {
  
  	          int n = a.length;
  
  	          Random r = new Random();
  
  	          for (int i = 0; i < n; i++) {
  	              int oi = r.nextInt(n);
  
  	              long temp = a[oi];
  	              a[oi] = a[i];
  	              a[i] = temp;
  	          }
  
  	          Arrays.sort(a);
  	      }
  
  		
  	    public static class MyScanner {
  	        BufferedReader br;
  	        StringTokenizer st;
  
  	        public MyScanner() {
  	            br = new BufferedReader(new InputStreamReader(System.in));
  	        }
  
  	        String next() {
  	            while (st == null || !st.hasMoreElements()) {
  	                try {
  	                    st = new StringTokenizer(br.readLine());
  	                } catch (IOException e) {
  	                    e.printStackTrace();
  	                }
  	            }
  	            return st.nextToken();
  	        }
  
  	        int nextInt() {
  	            return Integer.parseInt(next());
  	        }
  
  	        long nextLong() {
  	            return Long.parseLong(next());
  	        }
  
  	        double nextDouble() {
  	            return Double.parseDouble(next());
  	        }
  	        
  	        String nextLine() {
  	            String str = "";
  	            try {
  	                str = br.readLine();
  	            } catch (IOException e) {
  	                e.printStackTrace();
  	            }
  	            return str;
  	        }
  
  	    }
  	}
  