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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

//author:luckYrat(twitter:@luckYrat_)
//<ここに一言>

//せんげん！
#include <iostream>

#include <cmath>
#include <algorithm>
#include <iomanip>

#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <bitset>

#include <cctype>
#include <utility>
#include <climits>


//なまえくーかん！
using namespace std;
using ll = long long;
using P = pair<ll,ll>;

//てーすう！
const int mod = 1000000007;
const int inf = (1<<30)-1;
const ll linf = (1LL<<62LL)-1;
const double EPS = (1e-10);

//でふぁいん！
#define anyfill(n,s) setw(n) << setfill(s)
#define loop(s) for(int i = 0; s > i; i++)
#define rep(i,q) for(int i = 0; (q) > i; i++)
#define repp(i,n,q) for(int i = n; (q) > i; i++)
#define dep(i,q) for(int i = (q); 0 < i; i--)

//みじかく！
#define pb push_back
#define fir first
#define scn second
#define ednl endl

//いぇすのー！
#define YesNo(a) (a?"Yes":"No")
#define YESNO(a) (a?"YES":"NO")
#define yesno(a) (a?"yes":"no")

//きんぼーnほーこー！！
P ar4[4] = {{0,1},{0,-1},{1,0},{-1,0}};
P ar8[8] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
template <typename T>
T uPow(T z,T n, T mod){
  T ans = 1;
  while(n != 0){
    if(n%2){
      ans*=z;
      if(mod)ans%=mod;
    }
    n >>= 1;
    z*=z;
    if(mod)z%=mod;
  }
  return ans;
}
/*
確認ポイント
cout << fixed << setprecision(n) << 小数計算//n桁の小数表記になる

計算量は変わらないが楽できるシリーズ
min(max)_element(iter,iter)で一番小さい(大きい)値のポインタが帰ってくる
count(iter,iter,int)でintがiterからiterの間にいくつあったかを取得できる
*/


__attribute__((constructor))
void initial() {
 cin.tie(0);
 ios::sync_with_stdio(false);
}
int main(){
	queue<long long> A;
  for(int i = 1; 10 > i; i++){
    A.push(i);
  }
  int k;cin>>k;
  while(--k){
    long long z = A.front();A.pop();
    if(z%10!=0){
      A.push(z*10+(z%10)-1);
    }
    A.push(z*10+(z%10));
    if(z%10!=9){
      A.push(z*10+(z%10)+1);
    }
  }
  cout << A.front() << endl;
}