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 <algorithm>
#include <bitset>
#include <cassert>
#include <cfloat>
#include <cmath>
#include <climits>
#include <cstdlib>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <string>
#include <vector>
#define DEBUG 1
using namespace std;
constexpr int kMod = 1000000007;
typedef long long LL;

LL S(LL n) {
  LL ans = 0;
  while (n > 0) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}

int main() {
  LL K; cin >> K;
  set<LL> s;

  LL base = 1;
  for (LL i = 0; i < 15; ++i) {
    for (LL j = 1; j < 1000; ++j) {
      s.insert(base * (j + 1LL) - 1LL);
    }
    base *= 10LL;
  }

  vector<LL> v;
  double vmin = DBL_MAX;
  for (auto it = s.rbegin(); it != s.rend(); ++it) {
    LL x = *it;
    double y = (double)x / S(x);
    if (y <= vmin) {
      vmin = y;
      v.push_back((LL)x);
    }
  }

  reverse(v.begin(), v.end());
  for (int i = 0; i < K; ++i) cout << v[i] << endl;
}
