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;
typedef long long ll;
typedef pair<int,int> P;

ll next_snuke(ll in){
	ll dig = 1;
	ll ten = 1;

	double MIN = 1e20;
	ll ret;

	ll tmp = in;
	ll sum = 0;
	ll idx = 1;
	ll lis[20];


	while(tmp != 0){
		sum += tmp%10;
		tmp /= 10;
		lis[idx] = sum;
		idx++;
	}

	while(ten < in){
		ll top = (in/ten)%10;
		for (int i = top;i <= 9;i++){
			ll num = (in/(ten*10))*ten*10 + i*ten + 1*ten-1;
			ll snum = lis[idx-1] - lis[dig] + i + 9*(dig-1);
			double a = double(num)/double(snum);
			if(MIN > a){
				ret = num;
				MIN = a;
			}		
			//cout << num << " " << snum << " " << a << " from func \n";
		}
		dig++;
		ten *= 10;
	}

	return ret;
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	ll K; cin >> K;

	ll num = 1;
	cout << 1 << endl;

	for (int i = 1;i < K;i++){
		num = next_snuke(num+1);
		cout << num << endl;
	}

	return 0;

}
