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;
#define rep(i, n) for(long long i = 0; i < n; i++)
#define Rep(i, m, n) for(long long i = m; i < n; i++)
#define REP(i, m, n, p) for(long long i = m; i < n; i += p)
#define all(v) v.begin(), v.end()
#define pq priority_queue
#define bcnt(n) __builtin_popcountll(n)
using ll = long long;


int main() {
	// ll k;cin>>k;

	// rep(i,k){

	// 	ll g=(i+1)%9;
	// 	ll f=(i+1)/9;
	// 	if(g) cout<<g;
	// 	rep(i,f) cout<<9;
	// 	cout<<endl;
	// }

	ll p;
	cin >> p;

	ll d=1;//答えの最小間隔
	ll ans=1;
	Rep(j, 0, p) {
		ll n=ans;
		ll k=0;
		while(n){
			k+=n%10;
			n/=10;
		}
		ll l=0;
		n=ans+d;
		while(n){
			l+=n%10;
			n/=10;
		}
		if(ans*l<=(ans+d)*k) cout<<ans<<endl;
		else{
			j--;
			ans-=d;
			d*=10;
		}
		ans+=d;
	}
}