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
/*
g++ -std=c++11 -o2 -Wall ABC101D.cpp -o ./out/ABC101D.out
./out/ABC101D.out
*/

#include <bits/stdc++.h>
using namespace std;

#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define present(c, e) ((c).find((e)) != (c).end())
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define tr(c, i) for (iter(c) != (c).begin(); i != (c).end(); ++i)
#define pb push_back
#define mp make_pair

typedef long long ll;

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

int main(){
//    freopen("./input/input.txt", "r", stdin);
    ios::sync_with_stdio(0);
    cin.tie(0);
    ll k;
    cin >> k;
    ll n = 1, d = 1;
    cout << 1 << endl;
    for (ll i = 1; i < k; ++i)
    {
      if (sum(n + d) * (n + d * 10) < sum(n + d * 10) * (n + d))
        d *= 10;
      n += d;
      cout << n << endl;
    }

    return 0;
}
