Problem:
Score : 400 points

Problem Statement

A positive integer X is said to be a lunlun number if and only if the following condition is satisfied:

In the base ten representation of X (without leading zeros), for every pair of two adjacent digits, the absolute difference of those digits is at most 1.

For example, 1234, 1, and 334 are lunlun numbers, while none of 31415, 119, or 13579 is.

You are given a positive integer K. Find the K-th smallest lunlun number.

Constraints

1 \leq K \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

K

Output

Print the answer.

Sample Input 1

15

Sample Output 1

23

We will list the 15 smallest lunlun numbers in ascending order:

1,
2,
3,
4,
5,
6,
7,
8,
9,
10,
11,
12,
21,
22,
23.

Thus, the answer is 23.

Sample Input 2

1

Sample Output 2

1

Sample Input 3

13

Sample Output 3

21

Sample Input 4

100000

Sample Output 4

3234566667

Note that the answer may not fit into the 32-bit signed integer type.

Solution in Fortran (gfortran v4.8.4) (judge verdict: Accepted)

      MODULE solveMod
        implicit none
        
        public
        integer(16) :: counter=0,maxDigit,k
        integer(16) :: ans(10)
        
        contains
        function numLun( digit ) result( num )
          implicit none
          integer(16),intent(in) :: digit
          integer(16) :: num,i
          logical dummy
          
          maxDigit = digit
          
          do i = 1,9
            dummy = check( i,1_16 )
            ans(1) = i
          end do
          
          num = counter
        end function numLun
        
        recursive function check( num,numCall ) result( dummy )
          implicit none
          integer(16) :: num,numCall,i
          logical :: dummy
          character(10) :: s
          
          ans(numCall) = num
          
          if( num==-1.or.num==10 )then
            dummy = .false.
            return
          else
            if( numCall==maxDigit )then
              counter = counter + 1
              if(counter==k)then
                do i = 1,maxDigit
                  write(s(i:i),'(i1)') ans(i)
                end do
                !print*,ans(1:maxDigit)
                print*,s(1:maxDigit)
                stop
              end if
              return
            else
              dummy = check( num-1,numCall+1 )
              dummy = check( num  ,numCall+1 )
              dummy = check( num+1,numCall+1 )
            end if
          end if
        end function check
          
      END MODULE solveMod
      
      PROGRAM LunlunNumber
      use solveMod
      IMPLICIT NONE
      integer(16) ::  digit,buffer
      
      read*,k
      
      digit = 1 !digit is digit
      buffer = 0
      do while(.true.)
        buffer = buffer + numLun( digit )
        digit = digit + 1
      end do 
      
      
      END PROGRAM