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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.text.DecimalFormat;
  import java.util.*;
  
  //import sun.java2d.opengl.WGLSurfaceData.WGLVSyncOffScreenSurfaceData;
  
  
   
  public class Codeforces {
  	
  	static int mod=998244353  ;
  
  	public static void main(String[] args) throws Exception {
  		PrintWriter out=new PrintWriter(System.out);
  		FastScanner fs=new FastScanner();
  		int t=fs.nextInt();
  		while(t-->0) {
  			int n=fs.nextInt(), k=fs.nextInt();
  			int arr[]=fs.readArray(n);
  			Arrays.sort(arr);
  			long val[]=new long[n];
  			for(int i=0;i<n;i++) {
  				val[i]=(long)(Math.pow(10, arr[i]));
  			}
  			k++;
  			int i=0;
  			long res=0;
  			long cur=0;
  			while(k>0&&i<n-1) {
  				cur= Math.min(k, val[i+1]/val[i]-1);
  				res+= cur*val[i];
  				k-=cur;
  				i++;
  			}
  			res+= k*val[n-1];
  			System.out.println(res);
  		}
  		out.close();
  		
  	}
  	
  	static long pow(long a,long b) {
  		if(b<0) return 1;
  		long res=1;
  		while(b!=0) {
  			if((b&1)!=0) {
  				res*=a;
  				res%=mod;
  			}
  			a*=a;
  			a%=mod;
  			b=b>>1;
  		}
  		return res;
  	}
  	static long gcd(long  a,long  b) {
  		if(b==0) return a;
  		return gcd(b,a%b);
  	}
  	static long nck(int n,int k) {
  		if(k>n) return 0;
  		long res=1;
  		res*=fact(n);
  		res%=mod;
  		res*=modInv(fact(k));
  		res%=mod;
  		res*=modInv(fact(n-k)); 
  		res%=mod;
  		return res;
  	}
  	static long fact(long n) {
  //		return fact[(int)n];
  		long res=1;
  		for(int i=2;i<=n;i++) {
  			res*=i;
  			res%=mod;
  		}
  		return res;
  	}
  	
  	static long modInv(long n) {
  		return pow(n,mod-2);
  	}
  	
  	static void sort(int[] a) {
  		//suffle
  		int n=a.length;
  		Random r=new Random();
  		for (int i=0; i<a.length; i++) {
  			int oi=r.nextInt(n);
  			int temp=a[i];
  			a[i]=a[oi];
  			a[oi]=temp;
  		}
  		
  		//then sort
  		Arrays.sort(a);
  	}
  	
  	// Use this to input code since it is faster than a Scanner
  	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());
  		}
  		long[] lreadArray(int n) {
  			long a[]=new long[n];
  			for(int i=0;i<n;i++) a[i]=nextLong();
  			return a;
  		}
  		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());
  		}
  	}
   
  }