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 C {
  	
  	public static void main(String[] args)throws IOException {
  		
  		FastScanner scan = new FastScanner();
  		PrintWriter output = new PrintWriter(System.out);
  		int t = scan.nextInt();
  		for(int tt = 0;tt<t;tt++) {
  			int n = scan.nextInt(); 
  			long k = scan.nextLong();
  			long pow[] = new long[n];
  			for(int i = 0;i<n;i++) {
  				pow[i] = (long)Math.pow(10, scan.nextInt());
  			}
  			boolean found = false;
  			long notes = 0, cash = 0;
  			for(int i = 0;i<n-1;i++) {
  				notes+= pow[i+1]/pow[i]-1;
  				cash= pow[i+1]-1;
  				if(notes > k) {
  					cash-= (notes-k-1)*pow[i];
  					found = true;
  					break;
  				}
  			}
  			if(!found) {
  				cash+=(k-notes+1)*pow[n-1];
  			}
  			System.out.println(cash);
  			
  		}
  
  	}
  	
  	public static int[] sort(int arr[]) {
  		List<Integer> list = new ArrayList<>();
  		for(int i:arr) list.add(i);
  		Collections.sort(list);
  		for(int i = 0;i<list.size();i++) arr[i] = list.get(i);
  		return arr;
  	}
  	
  	public static int gcd(int a, int b) {
  		if(a == 0) return b;
  		return gcd(b%a, a);
  	}
  	
  	static boolean isPrime(int n) {
          if (n <= 1) return false;
          else if (n == 2) return true;
          else if (n % 2 == 0) return false;
          for (int i = 3; i <= Math.sqrt(n); i += 2) if (n % i == 0) return false;
          return true;
      }
  	
  	public static void printArray(int arr[]) {
  		for(int i:arr) System.out.print(i+" ");
  		System.out.println();
  	}
  	
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  
  }
  