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)

#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds; 

#define int long long
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define all(v) (v).begin(),(v).end()
#define f first
#define s second
#define sz(x) int(x.size())
#define endl '\n'
#define forn(i,n) for(int i=0;i<n;++i)
#define fore(i,l,r) for(int i=int(l);i<=int(r);++i)
#define rep(i,begin,end) for(__typeof(end) i=(begin);i!=(end);i++)
#define fill(a,value) memset(a,value,sizeof(a));
#define gcd(a,b) __gcd((a),(b))
#define watch1(x) cout << (x) << endl
#define watch2(x,y) cout << (x) << " " << (y) << endl
#define watch3(x,y,z) cout << (x) << " " << (y) << " " << (z) << endl
#define fastio ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
 
typedef long long ll;
typedef long double ld; 
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef tree<pii,null_type,less<pii>,rb_tree_tag,tree_order_statistics_node_update> oset;

// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;

const int mod = 998244353;
const int inf = 9e18;
const int N = 1e5 + 5;

void solve(){
    int i,k;
    cin>>k;
    vector<int> v;
    queue<int> q;
    for(i=1;i<10;++i){
        v.pb(i);
        q.push(i);
    }
    for(i=0;i<k;++i){
        int num=q.front();
        q.pop();
        int d=num%10;
        if(d!=0){
            v.pb(10*num+d-1);
            q.push(10*num+d-1);
        }
        v.pb(10*num+d);
        q.push(10*num+d);
        if(d!=9){
            v.pb(10*num+d+1);
            q.push(10*num+d+1);
        }
    }
    cout<<v[k-1];
}

signed main(){
    fastio;
    int t;
    //cin>>t;
    t=1;
    while(t--){
        solve();
    }
    return 0;
}