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 lint;
#define rep(i,n) for(lint i=0, i##_len=(n); i<i##_len; ++i)
#define all(x) (x).begin(),(x).end()
#define tsort(v) sort((v).begin(), (v).end())
#define trev(v) reverse((v).begin(), (v).end())
#define siz(x) ((int)(x).size())
#define el "\n"
#ifdef LOCAL_tqk
#define gc() getchar(); getchar()
#else
#define gc() 1
#endif

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }

double n_sn(lint n) {
	lint cp = n, s = 0;
	while (cp) {
		s += cp % 10;
		cp /= 10;
	}
	return (double)n / s;
}

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

	lint k, ans = 1, del = 1;
	cin >> k;
	rep(i, k) {
		cout << ans << el;
		if (n_sn(ans + del) > n_sn(ans + del * 10))del *= 10;
		//必ず9回ごとに del*=10 すると分母が100になるところでWAになる。
		//急にn_snが大きくなった時に del*=10 すれば解決...?
		ans += del;
	}
	gc();
}