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 Main {
  	
  	//----------- StringBuilder for faster output------------------------------
  	static StringBuilder out = new StringBuilder();
  
  
  	public static void main(String[] args) {
  		FastScanner fs=new FastScanner();
  
  		/******  CODE STARTS HERE  *****/
  		
  //		System.out.println((long)1e9);
  //		long d = (long)1e9+1 - 9*9;
  //		System.out.println(d);
  		long[] pre  = {1, 9, 99, 999, 9999, 99999, 999999, 9999999, 99999999, 999999999};
  		long[] pre_  = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000};
  		
  		int t = fs.nextInt();
  		while(t-->0) {
  			int n=fs.nextInt(), k=fs.nextInt();
  			int[] a = fs.readArray(n);
  			long sum = 0;
  			long res = k+1;
  			for(int i=1; i<n; i++) {
  				sum += pre[a[i]-a[i-1]];
  //				System.out.println(sum);
  				long temp = (k+1 - sum)*pre_[a[i]] + pre[a[i]];
  //				System.out.println(temp);
  				if(temp < 0)break;
  				res = temp;
  			}
  			out.append(res+"\n");
  		}
  		System.out.print(out);
  	}
      
  	static int countDigit(long n)
      {
          int count = 0;
          while (n != 0) {
              n = n / 10;
              ++count;
          }
          return count;
      }
  	
  	static void sort(int[] a) {
  		ArrayList<Integer> l=new ArrayList<>();
  		for (int i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
  	
  	
  	//----------- FastScanner class for faster input---------------------------
  	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());
  		}
  	}
  	
  }