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.math.BigInteger;
  import java.util.*;
  public class banknotes {
  
  	public static void main(String[] args) {
  		
  //		Scanner sc = new Scanner(System.in);
  		FastReader sc = new FastReader();
  		int t = sc.nextInt();
  		for(int o = 0 ; o<t;o++) {
  			int n = sc.nextInt();
  			int k = sc.nextInt();
  			k++;
  			int[] arr = new int[n];
  			for(int i = 0 ; i<n;i++) {
  				arr[i] = sc.nextInt();
  				}
  		long[] val = new long[n];
  		for(int i = 0 ; i<n;i++){
  			val[i] = (long)Math.pow(10, arr[i]);
  		}
  		long ans = 0;
  		String s1 = "";
  //		long yy = 9 + 90 + 999 + 9999 + 99999 + 999999+ 9999999 + 99999999;
  //		System.out.println(yy);
  //	long v = 999999920999999999l;
  //	System.out.println(v);
  		long [] dp = new long[n+1];
  		long rem = k;
  //		BigInteger bg = new BigInteger("0");
  		for(int i = 0 ; i<n-1;i++) {
  //			int x= i + 1;
  //			long temp = dp[i] + val[i] * (rem+1);
  //			
  //			//			BigInteger bg = new BigInteger("0");
  //				
  //			if(x == n) {
  //				ans = temp;
  //				break;
  //			}
  ////			System.out.println(temp);
  //			if(temp<=val[x] && temp>0) {
  //				ans = temp;
  //				break;
  //			}else{
  //				long num = val[i+1]-1;
  //				dp[x] = num*val[i] + dp[i];
  //				rem -= num;
  ////				System.out.println(dp[x]);
  ////			System.out.println(dp[x]);
  //			}
  //		
  		long v = (val[i + 1]/val[i]) - 1;
  		
  //		v = Math.min(v, rem);
  		v = Math.min(rem, v);
  		
  		ans += val[i]*v;
  		rem-=v;
  		if(rem== 0) {
  			break;
  		}
  		
  		}
  		if(rem>0) {
  			ans += (rem) * val[n-1];
  		}
  		System.out.println(ans);
  		
  		}
  	}
  
  }
  class FastReader {
      BufferedReader br;
      StringTokenizer st;
  
      public FastReader()
      {
          br = new BufferedReader(
              new InputStreamReader(System.in));
      }
  
      String next()
      {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
  
              
              
              
              
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
  
      int nextInt() { return Integer.parseInt(next()); }
  
      long nextLong() { return Long.parseLong(next()); }
  
      double nextDouble()
      {
          return Double.parseDouble(next());
      }
  
      String nextLine()
      {
          String str = "";
          try {
              str = br.readLine();
          }
          catch (IOException e) {
              e.printStackTrace();
          }
          return str;
      }
  }
  