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.

Example

Input

10


Output

1
2
3
4
5
6
7
8
9
19
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define For(i,a,b) for(i=a;i<b;i++)
#define rep(i,N) For(i,0,N)
#define rep1(i,N) For(i,1,N)

ll countdigits(ll n){
  ll ans=0;
  while(n){n/=10;ans++;}
  return ans;
}

ll sumdigits(ll n){
  ll ans=0;
  while(n){ans+=n%10;n/=10;}
  return ans;
}

ll f(ll n,vector<ll>& pow){
  ll i=0,j,digit=countdigits(n+1),tr=pow[digit-3]-1,ans=0;
  double p,min=(double)n/(double)sumdigits(n)*100.0;
  rep(i,1000){
    tr+=pow[digit-3];
    if(tr<=n)continue;
    p=(double)tr/(double)sumdigits(tr);
    if(min>p){
      min=p;
      ans=tr;
    }
  }
  return ans;
}

int main(){
  ll i,n=1099,K;
  vector<ll> pow(30,1);
  rep1(i,30)pow[i]=pow[i-1]*10;
  cin>>K;
  printf("1\n2\n3\n4\n5\n6\n7\n8\n9\n19\n");
  if(K==10)return 0;
  printf("29\n39\n49\n59\n69\n79\n89\n99\n199\n299\n399\n499\n599\n699\n799\n899\n999\n1099\n");
  K-=28;
  for(i=0;i<K;i++){
    n=f(n,pow);
    printf("%lld\n",n);
  }
  return 0;
}