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<bits/stdc++.h>
#define rep(i,a,b) for(register int i=a;i<=b;++i)
#define rpd(i,a,b) for(register int i=a;i>=b;--i)
#define rep1(i,x) for(register int i=head[x];i;i=nxt[i])
typedef long long ll;
const int N=1;
using namespace std;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m=1;
//int ji(int x){int ans=0;while(x)ans+=x%10,x/=10;return ans;}
int main(){
	n=read();
	//rep(i,1,n)if(i%10==9)printf("%.5lf %d\n",(double)i/(double)ji(i),i);
	while(n>9){rep(i,2,10)printf("%d\n",i*m-1);m*=10;n-=9;}
	rep(i,2,n+1)printf("%d\n",i*m-1);return 0;
}