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 PyPy3 (2.4.0) (judge verdict: Accepted)

import sys
sys.setrecursionlimit(10**6)
from math import floor,ceil,sqrt,factorial,log
from heapq import heappop, heappush, heappushpop
from collections import Counter,defaultdict,deque
from itertools import accumulate,permutations,combinations,product,combinations_with_replacement
from bisect import bisect_left,bisect_right
from copy import deepcopy
from operator import itemgetter
from fractions import gcd
mod = 10 ** 9 + 7
inf = float('inf')
ninf = -float('inf')
 
#整数input
def ii(): return int(sys.stdin.readline().rstrip()) #int(input())
def mii(): return map(int,sys.stdin.readline().rstrip().split())
def limii(): return list(mii()) #list(map(int,input().split()))
def lin(n:int): return [ii() for _ in range(n)]
def llint(n: int): return [limii() for _ in range(n)]
#文字列input
def ss(): return sys.stdin.readline().rstrip() #input()
def mss(): return sys.stdin.readline().rstrip().split()
def limss(): return list(mss()) #list(input().split())
def lst(n:int): return [ss() for _ in range(n)]
def llstr(n: int): return [limss() for _ in range(n)]

#本当に貪欲法か？ DP法では？？
#本当に貪欲法か？ DP法では？？
#本当に貪欲法か？ DP法では？？

k=ii()
ans=0
arr=deque([1,2,3,4,5,6,7,8,9])
while k>ans:
    n=int(arr.popleft())
    if n%10==9:
        arr.append(n*10 + n%10 -1)
        arr.append(n*10 + n%10 )
    elif n%10==0:
        arr.append(n*10 + n%10)
        arr.append(n*10 + n%10+1)
    else:
        arr.append(n*10 + n%10 -1)
        arr.append(n*10 + n%10)
        arr.append(n*10 + n%10 +1)
    ans+=1
print(n)