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 <vector>
#include <algorithm>
using namespace std;

long long getSumOfDigit(long long n) {
    return (n ? (n % 10) + getSumOfDigit(n / 10) : 0);
}

double getNDivSum(long long n) {
    return (double)n / getSumOfDigit(n);
}

int main() {
    long long k;
    cin >> k;

    vector<long long> ans;
    long long base = 1;

    for (int i = 0; i < 15; i++) {
        for (int j = 1; j < 9 * 15; j++) {
            ans.push_back((j + 1) * base - 1);
        }
        base *= 10;
    }
    sort(ans.begin(), ans.end());
    ans.erase(unique(ans.begin(), ans.end()), ans.end());

    for (long long i = 0; i < ans.size(); i++) {
        for (long long j = i + 1; j < ans.size(); j++) {
            if (getNDivSum(ans[i]) > getNDivSum(ans[j])) {
                ans.erase(ans.begin() + i);
                i--;
                break;
            }
        }
    }

    for (long long i = 0; i < k; i++) cout << ans[i] << endl;
}