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.*;
  import static java.lang.Math.*;
  
  public class C{
  	static  long[] p = new long[16];
  	static boolean[] w;
  	static long k;
  	
  	public static void main(String[] args) throws IOException {
  		// br = new BufferedReader(new FileReader(".in"));
  		// out = new PrintWriter(new FileWriter(".out"));
  		//new Thread(null, new (), "peepee", 1<<28).start();
  		
  		p[0]=1;
  		for (int  i =  1; i < p.length; i++) p[i]= p[i-1]*10;
  		// We know the maximum of sequence between two terms is 10^(j-i)-1
  		int  t =readInt();
  		
  		outer: while (t-->0) {
  			int n =readInt();
  			k =readLong();
  			w = new boolean[10];
  			while(n-->0) w[readInt()]=true;
  			// Build offa dat
  			long cur = 0;
  			int pr= 0;
  			long ans = 0;
  			for (int i =1; i< 10; i++) {
  				if (w[i]) {
  					long diff = p[i-pr]-1;
  					if (diff + cur > k) {
  						long need = k-cur+1;
  						//System.out.println(cur + " " + (k-cur));
  						//System.out.println("Ans  before " + ans);
  						ans += (need)*p[pr];
  						//System.out.println("Ans  after " + ans);
  						out.println(ans);
  						continue outer;
  					}
  					else {
  						cur += diff;
  						ans += (p[i-pr]-1)*p[pr];
  					}
  					pr = i;
  				}
  			}
  			long need = k-cur+1;
  			ans += need*p[pr];
  			out.println(ans);
  		}
  		
  		out.close();
  	}
  	
  	/* Stupid things to try if stuck:
  	 * n=1, expand bsearch range
  	 * brute force small patterns
  	 * submit stupid intuition
  	 */
  	
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  	static StringTokenizer st = new StringTokenizer("");
  	static String read() throws IOException{return st.hasMoreTokens() ? st.nextToken():(st = new StringTokenizer(br.readLine())).nextToken();}
  	static int readInt() throws IOException{return Integer.parseInt(read());}
  	static long readLong() throws IOException{return Long.parseLong(read());}
  	static double readDouble() throws IOException{return Double.parseDouble(read());}
  }