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.*;import java.util.*;import java.math.*;import static java.lang.Math.*;import static java.
          util.Map.*;import static java.util.Arrays.*;import static java.util.Collections.*;
  import static java.lang.System.*;
  public class Main
  {
      public void tq()throws Exception
      {
          st=new StringTokenizer(bq.readLine());
          int tq=i();
          sb=new StringBuilder(2000000);
          o:
          while(tq-->0)
          {
              int n=i();
              int k=i();
              long ar[]=arl(n);
              long c=0l;
              if(n==1)
              {
                  sl(k+1);
                  continue o;
              }
              for(int x=0;x<n;x++)
              {
                  long a=1l;
                  while(ar[x]-->0)a*=10;
                  ar[x]=a;
              }
              for(int x=1;x<n;x++)
              {
                  int v=(int)(ar[x]/ar[x-1]);
                  v--;
                  if(k<v)
                  {
                      c+=ar[x-1]*(k+1);
                      sl(c);
                      continue o;
                  }
                  else
                  {
                      c+=ar[x-1]*v;
                      k-=v;
                  }
              }
              c+=ar[n-1]*(k+1);
              sl(c);
          }
          p(sb);
      }
      void f(){out.flush();}
      int di[][]={{-1,0},{1,0},{0,-1},{0,1}};
      int de[][]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{1,1},{-1,1},{1,-1}};
      long mod=1000000007l;int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;long maxl=Long.MAX_VALUE, minl=Long.
          MIN_VALUE;BufferedReader bq=new BufferedReader(new InputStreamReader(in));StringTokenizer st;
      StringBuilder sb;public static void main(String[] a)throws Exception{new Main().tq();}int[] so(int ar[])
  {Integer r[]=new Integer[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];sort(r);for(int x=0;x<ar.length;
                                                                                         x++)ar[x]=r[x];return ar;}long[] so(long ar[]){Long r[]=new Long[ar.length];for(int x=0;x<ar.length;x++)
      r[x]=ar[x];sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
      char[] so(char ar[]) {Character
              r[]=new Character[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];sort(r);for(int x=0;x<ar.length;x++)
          ar[x]=r[x];return ar;}void s(String s){sb.append(s);}void s(int s){sb.append(s);}void s(long s){sb.
          append(s);}void s(char s){sb.append(s);}void s(double s){sb.append(s);}void ss(){sb.append(' ');}void sl
          (String s){sb.append(s);sb.append("\n");}void sl(int s){sb.append(s);sb.append("\n");}void sl(long s){sb
          .append(s);sb.append("\n");}void sl(char s) {sb.append(s);sb.append("\n");}void sl(double s){sb.append(s)
  ;sb.append("\n");}void sl(){sb.append("\n");}int l(int v){return 31-Integer.numberOfLeadingZeros(v);}
      long l(long v){return 63-Long.numberOfLeadingZeros(v);}int sq(int a){return (int)sqrt(a);}long sq(long a)
  {return (long)sqrt(a);}long gcd(long a,long b){while(b>0l){long c=a%b;a=b;b=c;}return a;}int gcd(int a,int b)
  {while(b>0){int c=a%b;a=b;b=c;}return a;}boolean pa(String s,int i,int j){while(i<j)if(s.charAt(i++)!=
          s.charAt(j--))return false;return true;}boolean[] si(int n) {boolean bo[]=new boolean[n+1];bo[0]=true;bo[1]
          =true;for(int x=4;x<=n;x+=2)bo[x]=true;for(int x=3;x*x<=n;x+=2){if(!bo[x]){int vv=(x<<1);for(int y=x*x;y<=n;
                                                                                                       y+=vv)bo[y]=true;}}return bo;}long mul(long a,long b,long m) {long r=1l;a%=m;while(b>0){if((b&1)==1)
      r=(r*a)%m;b>>=1;a=(a*a)%m;}return r;}int i()throws IOException{if(!st.hasMoreTokens())st=new
          StringTokenizer(bq.readLine());return Integer.parseInt(st.nextToken());}long l()throws IOException
  {if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return Long.parseLong(st.nextToken());}String
  s()throws IOException {if (!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return st.nextToken();}
      double d()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return Double.
              parseDouble(st.nextToken());}void p(Object p){out.print(p);}void p(String p){out.print(p);}void p(int p)
  {out.print(p);}void p(double p){out.print(p);}void p(long p){out.print(p);}void p(char p){out.print(p);}void
  p(boolean p){out.print(p);}void pl(Object p){out.println(p);}void pl(String p){out.println(p);}void pl(int p)
  {out.println(p);}void pl(char p){out.println(p);}void pl(double p){out.println(p);}void pl(long p){out.
          println(p);}void pl(boolean p)
  {out.println(p);}void pl(){out.println();}void s(int a[]){for(int e:a)
  {sb.append(e);sb.append(' ');}sb.append("\n");}
      void s(long a[])
      {for(long e:a){sb.append(e);sb.append(' ')
      ;}sb.append("\n");}void s(int ar[][]){for(int a[]:ar){for(int e:a){sb.append(e);sb.append(' ');}sb.append
          ("\n");}}
      void s(char a[])
      {for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}void s(char ar[][])
  {for(char a[]:ar){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}int[] ari(int n)throws
          IOException {int ar[]=new int[n];if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());for(int x=0;
                                                                                                            x<n;x++)ar[x]=Integer.parseInt(st.nextToken());return ar;}int[][] ari(int n,int m)throws
          IOException {int ar[][]=new int[n][m];for(int x=0;x<n;x++){if (!st.hasMoreTokens())st=new StringTokenizer
          (bq.readLine());for(int y=0;y<m;y++)ar[x][y]=Integer.parseInt(st.nextToken());}return ar;}long[] arl
          (int n)throws IOException {long ar[]=new long[n];if(!st.hasMoreTokens()) st=new StringTokenizer(bq.readLine())
          ;for(int x=0;x<n;x++)ar[x]=Long.parseLong(st.nextToken());return ar;}long[][] arl(int n,int m)throws
          IOException {long ar[][]=new long[n][m];for(int x=0;x<n;x++) {if(!st.hasMoreTokens()) st=new
          StringTokenizer(bq.readLine());for(int y=0;y<m;y++)ar[x][y]=Long.parseLong(st.nextToken());}return ar;}
      String[] ars(int n)throws IOException {String ar[] =new String[n];if(!st.hasMoreTokens())st=new
              StringTokenizer(bq.readLine());for(int x=0;x<n;x++)ar[x]=st.nextToken();return ar;}double[] ard
          (int n)throws IOException {double ar[] =new double[n];if(!st.hasMoreTokens())st=new StringTokenizer
          (bq.readLine());for(int x=0;x<n;x++)ar[x]=Double.parseDouble(st.nextToken());return ar;}double[][] ard
          (int n,int m)throws IOException{double ar[][]=new double[n][m];for(int x=0;x<n;x++) {if(!st.hasMoreTokens())
      st=new StringTokenizer(bq.readLine());for(int y=0;y<m;y++) ar[x][y]=Double.parseDouble(st.nextToken());}
      return ar;}char[] arc(int n)throws IOException{char ar[]=new char[n];if(!st.hasMoreTokens())st=new
          StringTokenizer(bq.readLine());for(int x=0;x<n;x++)ar[x]=st.nextToken().charAt(0);return ar;}char[][]
  arc(int n,int m)throws IOException {char ar[][]=new char[n][m];for(int x=0;x<n;x++){String s=bq.readLine();
      for(int y=0;y<m;y++)ar[x][y]=s.charAt(y);}return ar;}void p(int ar[])
  {StringBuilder sb=new StringBuilder
          (2*ar.length);for(int a:ar){sb.append(a);sb.append(' ');}out.println(sb);}void p(int ar[][])
  {StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);for(int a[]:ar){for(int aa:a){sb.append(aa);
      sb.append(' ');}sb.append("\n");}p(sb);}void p(long ar[]){StringBuilder sb=new StringBuilder
          (2*ar.length);for(long a:ar){ sb.append(a);sb.append(' ');}out.println(sb);}
      void p(long ar[][])
      {StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);for(long a[]:ar){for(long aa:a){sb.append(aa);
          sb.append(' ');}sb.append("\n");}p(sb);}void p(String ar[]){int c=0;for(String s:ar)c+=s.length()+1;
      StringBuilder sb=new StringBuilder(c);for(String a:ar){sb.append(a);sb.append(' ');}out.println(sb);}
      void p(double ar[])
      {StringBuilder sb=new StringBuilder(2*ar.length);for(double a:ar){sb.append(a);
          sb.append(' ');}out.println(sb);}void p
          (double ar[][]){StringBuilder sb=new StringBuilder(2*
          ar.length*ar[0].length);for(double a[]:ar){for(double aa:a){sb.append(aa);sb.append(' ');}sb.append("\n")
  ;}p(sb);}void p(char ar[])
  {StringBuilder sb=new StringBuilder(2*ar.length);for(char aa:ar){sb.append(aa);
      sb.append(' ');}out.println(sb);}void p(char ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0]
          .length);for(char a[]:ar){for(char aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}void pl
          (int... ar){for(int e:ar)p(e+" ");pl();}void pl(long... ar){for(long e:ar)p(e+" ");pl();}
  }