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;
using ld = long double;

const int INF = (1 << 30) - 1;
const ll INF64 = ((ll)1 << 62) - 1;
const double PI = 3.1415926535897932384626433832795;

const int dx[] = { 0, 1, 0, -1 };
const int dy[] = { -1, 0, 1, 0 };

ll f(ll n) {
	ll res = 0;
	while (n > 0) {
		res += n % 10;
		n /= 10;
	}

	return res;
}

double g(ll n) {
	return (double)(n) / f(n);
}

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

	vector<ll> res;
	ll base = 1;

	for (int i = 0; i < 15; i++) {
		for (int j = 1; j < 150; j++) {
			res.emplace_back(base * (j + 1) - 1);
		}
		base *= 10;
	}

	sort(begin(res), end(res));
	res.erase(unique(begin(res), end(res)), end(res));

	for (ll i = 0; i < res.size(); i++) {
		for (ll j = i + 1; j < res.size(); j++) {
			if (g(res[i]) > g(res[j])) {
				res.erase(begin(res) + i--);
				break;
			}
		}
	}

	ll k;
	cin >> k;

	for (ll i = 0; i < k; i++) {
		cout << res[i] << endl;
	}
	return 0;
}
