Problem:
Score : 500 points

Problem Statement

Let S(n) denote the sum of the digits in the decimal notation of n.
For example, S(123) = 1 + 2 + 3 = 6.

We will call an integer n a Snuke number when, for all positive integers m such that m > n, \frac{n}{S(n)} \leq \frac{m}{S(m)} holds.

Given an integer K, list the K smallest Snuke numbers.

Constraints

1 \leq K

The K-th smallest Snuke number is not greater than 10^{15}.

Input

Input is given from Standard Input in the following format:

K

Output

Print K lines. The i-th line should contain the i-th smallest Snuke number.

Sample Input 1

10

Sample Output 1

1
2
3
4
5
6
7
8
9
19

Solution in C++14 (Clang 3.8.0) (judge verdict: Runtime Error)

//#include "bits/stdc++.h"
#include <iostream>
#include <vector>
#include <map>
#include <utility>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <numeric>
#include <complex>
#include <bitset>
#include <functional>
#include <stack>
#include <regex>
#include <tuple>
#include <iomanip>

#define int long long
#define REP(i,a,n) for(int i=a;i<n;++i)
#define rep(i,n) REP(i,0,n)
#define REV(i,a,n) for(int i=n;i>=a;--i)
#define all(e) e.begin(),e.end()
#define rall(e) e.rbegin(),e.rend()
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define show(n) cerr<<#n<<" = "<<n<<endl
#define showp(n) cerr<<n.fs<<", "<<n.sc<<endl
#define shows(n) for(auto z:n){cerr<<z<<", ";}cerr<<endl
#define showsp(n) for(auto z:n){cerr<<z.fs<<" "<<z.sc<<", "}cerr<<endl

#define yes puts("Yes")
#define no puts("No")
#define case(i) printf("Case #%lld: ",i)

using namespace std;

using vi=vector<int>;
using pint=pair<int,int>;

inline void io(){cin.tie(0);ios::sync_with_stdio(false);cout.tie(0);cout<<fixed<<setprecision(20);}
inline int  in(){int n;scanf("%lld",&n);return n;}
inline void out(int n){printf("%lld\n",n);}

const int INF=1LL<<55;
const int MOD=1000000007;
const double EPS=1e-8;

int k;

int digitsum(int n){
    int sum=0;
    while(n>0){
        sum+=n%10;
        n/=10;
    }
    return sum;
}

signed main(){
    io();

    cin>>k;
    int cur=1;

    int plus=1;
    rep(i,k){
        printf("%lld\n", cur);
        
        cur+=plus;
        if((cur+1LL)%(plus*10LL)==0LL) plus*=10LL;
    }

    return 0;
}