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>
#include <iomanip>
using namespace std;
#define reps(i,s,n) for(int i = s; i < n; i++)
#define rep(i,n) reps(i,0,n)
#define fi first
#define se second
#define mp make_pair
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

ll N,M,H,W,K,Q,A,B,L,R;
string S,T;
const ll MOD = (1e+9) + 7;
const ll INF = 1LL << 60;
typedef pair<ll,ll> P;

typedef vector<P> vp;
typedef vector<vp> matP;

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

int main(){
    cin>>K;
    N = 1;
    vec bit10(19,1);
    reps(i,1,19) bit10[i] = bit10[i-1] * 10;
    rep(i,K){
        cout<<N<<endl;
        ++N;
        ll next = N;
        for(ll m = N, d = 0; m >= 0; m /= 10){
            ll temp = bit10[d+1] * (N/bit10[d+1] + 1) - 1;
            if(next * sn(temp) > temp * sn(next)) next = temp;
            ++d;
            if(m == 0) break;
            //cout<<i<<' '<<m<<endl;
        }
        N = next;
    }
}
