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.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.StringTokenizer;
  
  public class Solution{
  	
  	static int max(int[] arr,int n){
  		int mx = Integer.MIN_VALUE;
  		for(int i = 0;i < n;i++){
  			if(arr[i] > mx){
  				mx = arr[i];
  			}
  		}
  		return mx;
  	}
  
  	static int min(int[] arr,int n){
  		int mn = Integer.MAX_VALUE;
  		for(int i = 0;i < n;i++){
  			if(arr[i] < mn){
  				mn = arr[i];
  			}
  		}
  		return mn;
  	}
  
  	static long binexp(long a,long b){
  		long ans = 1;
  		while(b > 0){
  			if((b & 1l) == 1l) ans *= a;
  			a *= a;
  			b >>= 1;
  		}
  		return ans;
  	}
  
  	static void sort(int[] arr) {
  		ArrayList<Integer> l = new ArrayList<>();
  		for (int i:arr) l.add(i);
  		Collections.sort(l);
  		for (int i = 0; i < arr.length; i++) arr[i] = l.get(i);
  	}
  
  	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 nextLong(){
  			return Long.parseLong(next());
  		}
  		float nextFloat(){
  			return Float.parseFloat(next());
  		}
  		double nextDouble(){
  			return Double.parseDouble(next());
  		}
  		int[] int_arr(int n){
  			int arr[] = new int[n];
  			for(int i = 0;i < n;i++) arr[i] = nextInt();
  			return arr;
  		}
  		long[] l_arr(int n){
  			long arr[] = new long[n];
  			for(int i = 0;i < n;i++) arr[i] = nextLong();
  			return arr;
  		}
  	}
  
  	static int bound(int[] arr,int start,int end,int x,boolean lower){
  		int ans = -1;
  		while(start <= end){
  			int mid = start + (end - start) / 2;
  			if(arr[mid] > x) end = mid - 1;
  			else if(arr[mid] < x) start = mid + 1;
  			else{
  				ans = mid;
  				if(lower) end = mid - 1;
  				else start = mid + 1;
  			}
  		}
  		return ans;
  	}
  
  	static boolean isPrime(int n){
          if(n == 1){
              return false;
          }
          if(n == 2 || n == 3){
              return true;
          }
          if(n % 2 == 0 || n % 3 == 0){
              return false;
          }
          for(long i = 5; i*i <= n; i += 6){
              if(n % i == 0 || n % (i + 2) == 0)
                  return false;
          }
          return true;
      }
  
  	static int gcd(int a,int b){
  		return (a == 0) ? b : gcd(b % a,a);
  	}
  
  	static int l = 10000000;
  	static boolean[] sieve = new boolean[l+1];
  	static void primes(){
  		Arrays.fill(sieve,true);
  		for(int i = 2;i * i <= l;i++){
  			if(sieve[i]){
  				for(int j = i * i;j <= l;j += i){
  					sieve[j] = false;
  				}
  			}
  		}
  	}
  
  	public int bs(int[] arr,int start,int end,int target){
  		while(start <= end){
  			int mid = start + (end - start) / 2;
  			if(arr[mid] == target){
  				return mid;
  			}
  			else if(arr[mid] > target){
  				end = mid - 1;
  			}
  			else{
  				start = mid + 1;
  			}
  		}
  		return -1;
  	}
  
  	public static void main(String[] args){
  		FastScanner in = new FastScanner();
  		PrintWriter out = new PrintWriter(System.out);
  
  		int t = in.nextInt();
  		while(t --> 0){
  			long n = in.nextLong(),k = in.nextLong();
  			long[] arr = new long[(int)n];
  			for(int i = 0;i < (int)n;i++) arr[i] = binexp(10l,in.nextLong());
  			long tot = 0;
  			k += 1;
  			for(int i = 1;i < (int)n;i++){
  				long val = (arr[i] / arr[i - 1]) - 1;
  				if(val < k){
  					tot += val * arr[i - 1];
  				}
  				else{
  					tot += k * arr[i - 1];
  				}
  				k -= val;
  				if(k <= 0) break;
  			}
  			if(k > 0){
  				tot += k * arr[(int)n - 1];
  			}
  			out.println(tot);
  		}
  		out.close();
  		
      }
  }
  