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 <iomanip>
#include <vector>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <cstring>
#define rep(i, N) for (int i = 0; i < (int)N; i++)
using namespace std;
typedef long long ll;
const ll LLINF = 9223372036854775807;
const int MOD = 1000000007;

int dsum(ll n) {
  int result = 0;
  while (n!=0) {result+=n%10; n/=10;}
  return result;
}

int main() {
  int K; cin >> K;
  rep(i,min(K,9)) { cout << i+1 << endl; }
  if (K<=9) return 0;
  K -= 9;
  ll result = 9, digsum = 9;
  rep(i,K) {
    ll r = result, d = digsum;
    for (int dig = 0; pow(10,dig)<=r; dig++) {
      ll tr = ceil(r/pow(10,dig+1)+1)*pow(10,dig+1)-1;
      ll td = 0; for (ll tmp = tr; tmp>0; tmp/=10) td += tmp%10;
      if (dig==0||result*td>=tr*digsum) { result = tr; digsum = td; }
      // cout << tr << " " << td << endl;
    }
    cout << result << endl;
  }
  return 0;
}