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 (GCC 5.4.1) (judge verdict: Wrong Answer)

# include <iostream>
# include <algorithm>
#include <array>
# include <cassert>
#include <cctype>
#include <climits>
#include <numeric>
# include <vector>
# include <string>
# include <set>
# include <map>
# include <cmath>
# include <iomanip>
# include <functional>
# include <tuple>
# include <utility>
# include <stack>
# include <queue>
# include <list>
# include <bitset>
# include <complex>
# include <chrono>
# include <random>
# include <limits.h>
# include <unordered_map>
# include <unordered_set>
# include <deque>
# include <cstdio>
# include <cstring>
#include <stdio.h>
#include<time.h>
#include <stdlib.h>
#include <cstdint>
#include <cfenv>
//#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using ULL = unsigned long long;
long long MOD = 1000000000+7;
constexpr long long INF = numeric_limits<LL>::max();
const double PI = acos(-1);
#define fir first
#define sec second
#define thi third
#define debug(x) cerr<<#x<<": "<<x<<'\n'
typedef pair<LL, LL> Pll;
typedef pair<LL, pair<LL, LL>> Ppll;
typedef pair<LL, pair<LL, bitset<100001>>> Pbll;
typedef pair<LL, pair<LL, vector<LL>>> Pvll;
typedef pair<LL, LL> Vec2;
struct Tll { LL first, second, third; };
typedef pair<LL, Tll> Ptll;
#define rep(i,rept) for(LL i=0;i<rept;i++)
#define Mfor(i,mf) for(LL i=mf-1;i>=0;i--)
LL h, w, n, m, k, t, s, q, last, cnt, sum[2000000], ans, d[200000], a[2000000], b[2000000],rnd[200000];
string str[2],ss;
bool f;
char c;
int di[4][2] = { { 0,1 },{ 1,0 },{ -1,0 } ,{ 0,-1 } };
struct Edge { LL to, cost; };
vector<LL>vec[100000],hashs[100000];
vector<Pll>v;
map<string, LL>ma;
set<LL>st;
void YN(bool f) {
	if (f)
		cout << "YES" << endl;
	else
		cout << "NO" << endl;
}
void yn(bool f) {
	if (f)
		cout << "Yes" << endl;
	else
		cout << "No" << endl;
}
LL func(LL x) {
	LL ret = 0;
	while (x > 0) {
		ret += x % 10;
		x /= 10;
	}
	return ret;
}
pair<double, LL>pd[2000000];
int main() {
	cin >> n;
	q = 1;
	rep(i,10) {
		rep(j, 100000) {
			double out = (j + 1) *q - 1;
			if (func(out) != 0) {
				pd[i * 100000 + j].sec = out;
				pd[i * 100000 + j].fir = out / func(out);
			}
		}
		q *= 10;
	}
	sort(pd, pd + 2000000);
	LL i = 0;
	while (1) {
		if (last<pd[i].second) {
			cout << pd[i].second << endl;
			last = pd[i].second;
			cnt++;
			if (cnt == 693)
				break;
			if (cnt == n)
				return 0;
		}
		i++;
	}
	rep(i, n - cnt) {
		cout << 1 << endl;
	}
		return 0;
}