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 <bits/stdc++.h>
using namespace std;
#define ll long long int
vector<ll> tostr(ll n){
    vector<ll> num;
    while(n){
        num.push_back(n%10);
        n/=10;
    }
    return num;
}
double f(ll n, ll sum){
    return (double)n/(double)sum;
}
ll nextSnuke(vector<ll> num, ll n){
    ll sum=0;
    for(ll a=0;a<num.size();a++){
        sum+=num[a];
    }
    double mini=f(n,sum);
    ll sepuluh=1;
    ll ans=n;//printf("%d %d %lf\n",n,sum,mini);
    for(ll a=0;a<num.size();a++){
        n=n-num[a]*sepuluh+9*sepuluh;
        sum=sum-num[a]+9;
        double tmp=f(n,sum);//printf("%d %d %d %lf\n",num[a],n,sum,tmp);
        if(tmp<mini){
            mini=tmp;
            ans=n;
        }
        sepuluh*=10;
    }
    return ans;
}
int main(){
    ll k;
    scanf("%lld",&k);
    ll snuke=1;
    printf("%d\n",snuke);
    k--;
    while(k--){
        vector<ll> num = tostr(snuke+1);
        snuke = nextSnuke(num,snuke+1);
        printf("%lld\n",snuke);
    }
}
