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.Scanner;
  
  public class Main {
      public static final int [] nums = {1, 10, 100,1000,10000,100000,1000000,10000000,100000000,1000000000};
      public static long min(long a, long b) {
          return a>b?b:a;
      }
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int t = sc.nextInt();
          while (t-- > 0) {
              int n = sc.nextInt();
              long k = sc.nextLong() + 1;
              int []a = new int[n];
              int i=-1;
              while(++i<n) a[i] = sc.nextInt();
              long ans = 0;
              for(int j=0; j < n-1&&k>0; j++) {
                  long v = min(nums[a[j+1]]-nums[a[j]], k*nums[a[j]]);
                  ans += v;
                  k -= v / nums[a[j]];
              }
              ans += k *nums[a[n-1]];
              System.out.println(ans);
  
              // int ans = 0;
              // long x = 1;
              // while (x <= k && x < n) {
              //     ans += 1;
              //     if (n - x >= x) {
              //         x += x;
              //     } else {
              //         x += n - x;
              //     }
              // }
              // if(n > x) {
              //     ans += (n-x+k-1)/k;
              // }
              // System.out.println(ans);
          }
          sc.close();
      }
  }