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 C{
  	public static void main(String[] args){
  		FastReader sc = new FastReader();
  		int t=sc.nextInt();
  		while(t-->0){
  			int n=sc.nextInt();
  			int k=sc.nextInt()+1;
  			int a[]=sc.fastArray(n);
  			int b[]=a.clone();
  			for(int i=0;i<n;i++)
  				a[i]=(int)Math.pow(10, a[i]);
  			int max[]=new int [n];
  			if(n>1)max[0]=a[1]-1;
  			max[n-1]=9;
  			for(int i=1;i<n-1;i++) {
  				if(b[i+1]-1!=b[i]) {
  					int zero=b[i+1]-b[i];
  					int num=1;
  					while(zero>0) {
  						num*=10;
  						zero--;
  					}
  					max[i]=num-1;
  						
  				}else max[i]=9;
  			}
  			long num=0L;int j=0;
  			while(k>0 && j<n-1) {
  				int x=Math.min(k, max[j]);
  				num+=((long)x*(long)a[j]);
  				j++;
  				k-=x;
  			}
  			if(k>0) num+=(((long)k*(long)a[n-1]));
  			System.out.println(num);
  		}		 
      }
  
  	static class pair {
  		int x;int y;
  		 pair(int x,int y){
  			this.x=x;
  			this.y=y;
  		}
  		
  	}
  	static ArrayList<Integer> primeFac(int n){
  		ArrayList<Integer>ans = new ArrayList<Integer>();
  		int lp[]=new int [n+1];
  		Arrays.fill(lp, 0);  	//0-prime
  		for(int i=2;i<=n;i++) {
  			if(lp[i]==0) {
  				for(int j=i;j<=n;j+=i) {
  					if(lp[j]==0) lp[j]=i;
  				}
  			}
  		}
  		int fac=n;
  		while(fac>1) {
  			ans.add(lp[fac]);
  			fac=fac/lp[fac];
  		}
  		print(ans);
  		return ans;
  	}
  	static ArrayList<Long> prime_in_given_range(long l,long r){
  		ArrayList<Long> ans= new ArrayList<>();
  		int n=(int)Math.sqrt(r)+1;
  		int prime[]=sieve_of_Eratosthenes(n);
  		long res[]=new long [(int)(r-l)+1];
  		for(int i=0;i<=r-l;i++) {
  			res[i]=i+l;
  		}
  		for(int i=0;i<prime.length;i++) {
  			if(prime[i]==1) {
  				System.out.println(2);
  				for(int j=Math.max((int)i*i, (int)(l+i-1)/i*i);j<=r;j+=i) {
  					res[j-(int)l]=0; 
  				}
  			}
  		}
  		for(long i:res) if(i!=0)ans.add(i);
  		return ans;
  	}
  	static int [] sieve_of_Eratosthenes(int n) {
  		int prime[]=new int [n];
  		Arrays.fill(prime, 1);  // 1-prime | 0-not prime
  		prime[0]=prime[1]=0;
  		for(int i=2;i<n;i++) {
  			if(prime[i]==1) {
  				for(int j=i*i;j<n;j+=i) {
  					prime[j]=0;
  				}
  			}
  		}
  		return prime;
  	}
  	static long binpow(long a,long b) {
  		long res=1;
  		if(b==0)return a;
  		if(a==0)return 1;
  		while(b>0) {
  			if((b&1)==1) {
  				res*=a;
  			}
  			a*=a;
  			b>>=1;
  		}
  		return res;
  	}
  	static void print(int a[]) {
  		System.out.println(a.length);
  		for(int i:a) {
  			System.out.print(i+" ");
  		}
  		System.out.println();
  	}
  	static void print(long a[]) {
  		System.out.println(a.length);
  		for(long i:a) {
  			System.out.print(i+" ");
  		}
  		System.out.println();
  	}
  	static void print(ArrayList<Integer> a) {
  		System.out.println(a.size());
  		for(long i:a) {
  			System.out.print(i+" ");
  		}
  		System.out.println();
  	}
  	static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader()
  		{
  			br = new BufferedReader(
  				new InputStreamReader(System.in));
  		}
  		int [] fastArray(int n) {
  			int a[]=new int [n];
  			for(int i=0;i<n;i++) {
  				a[i]=nextInt();
  			}
  			return a;
  		}
  		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;
  		}
  	}
  
  }