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 <iostream>
#include<vector>
#include<algorithm>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;

const ll MOD_CONST = 1000000007;

ll digit10(ll num) {
	ll d = 0;
	while (num > 0) {
		d++;
		num /= 10;
	}
	return d;
}

ll s(ll x) {
	int sum = 0;
	while (x > 0) {
		sum += (x % 10);
		x /= 10;
	
	}
	return sum;
}

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

	
	ll num = 1;
	
	for (ll i = 0; i < k; i++) {
		cout << num << endl;
		num++;
		priority_queue<ll, vector<ll>, greater<ll>> pq;
		pq.push(num);
		ll d_max = digit10(num) + 1;
		ll p10 = 10;
		for (int i = 0; i <= d_max; i++) {
			
			pq.push((num / p10 + 1) * p10 - 1);
			p10 *= 10;
		}

		ll minX = pq.top();
		pq.pop();
		ll minS = s(minX);

		for (int i = 0; i < d_max; i++) {
			ll x1 = pq.top();
			ll s1 = s(x1);
			pq.pop();
			//cout << x1 << endl;
			if (x1 * minS < minX * s1) {
				minX = x1;
				minS = s1;
			}

		}
		

		num = minX;
		
		
	}
}  

