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<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<deque>
#include<map>
#include<queue>
using namespace std;
#define rep(i,s,n) for(int i = s;i<n;i++)
#define repe(i,s,n) for(int i = s;i<=n;i++)
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef pair<int,int>pint;
typedef vector<int>vint;
typedef vector<pint>vpint;
static const ll maxLL = (ll)1 << 62;
const int mod=1000000007;
int dy[]={-1,0,1,0};
int dx[]={0,1,0,-1};

/*#define int ll*/
int sum(long long n) {
    int r = 0;
    while (n > 0) {
        r += n % 10;
        n /= 10;
    }
    return r;
}
int main(){
    int k;
    cin >> k;
    long long a=1;
    long long p=1;
    for(int i=1;i<=k;i++){
        cout << a<< endl;
        long long c=a+p;
        long long d=a+p*10;
        if(c* sum(d) <= d * sum(c)){
            a=c;
        }
        else {
            a=d;
            p=p*10;
        }
    }
}