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.util.*;
  import java.lang.*;
  import java.io.*;
  
  public class Main
  {
  	PrintWriter out = new PrintWriter(System.out);
  	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
      StringTokenizer tok = new StringTokenizer("");
      String next() throws IOException {
          if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
          return tok.nextToken();
      }
      int ni() throws IOException { return Integer.parseInt(next()); }
      long nl() throws IOException { return Long.parseLong(next()); }
      
      long mod=1000000007;
      
      void solve() throws IOException {
          for (int tc=ni();tc>0;tc--) {
              int n=ni();
              long k=nl();
              long[]A=new long[n];
              for (int i=0;i<n;i++) {
                  A[i]=1;
                  int pow=ni();
                  for (int j=0;j<pow;j++) A[i]*=10;
              }
              
              long ans=0;
              for (int i=1;i<n;i++) {
                  long need=A[i]/A[i-1]-1;
                  if (need>k) { ans+=(k+1)*A[i-1]; k=-1; break; }
                  ans+=need*A[i-1];
                  k-=need;
                  //out.println(ans+" "+k);
              }
              
              ans+=(k+1)*(A[n-1]);
              
              out.println(ans);
          }
          out.flush();
      }
      
      int gcd(int a,int b) { return(b==0?a:gcd(b,a%b)); }
      long gcd(long a,long b) { return(b==0?a:gcd(b,a%b)); }
      long mp(long a,long p) { long r=1; while(p>0) { if ((p&1)==1) r=(r*a)%mod; p>>=1; a=(a*a)%mod; } return r; }
      
      public static void main(String[] args) throws IOException {
          new Main().solve();
      }
  }