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>
#define rep(i,n) for(int i = 0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;
//long long ccw(const Point& A, const Point& B, const Point& C) {
//	return 1LL * (B.x - A.x)*(C.y - A.y) - 1LL * (B.y - A.y)*(C.x - A.x);
//}

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

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

int main() {
	// 최대유량, 앳코더, (종만북(?) 생각) - 기하 잘하고 싶다!
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	// 1 2 3 4 5 6 7 8 9 19 29 39 49 59 69 79 89 99 109 ... 
	ll k;
	cin >> k;
	ll n = 1, a = 1;
	while (k--) {
		cout << n << endl;
		while (f(n + a) > f(n + 2 * a)) {
			a *= 10;
		}// 99 + 10 , 99 + 20  10.9 , 119 // 11
		n += a;
	}

}
