Problem: There are n digital panels placed in a straight line. Each panel can show any digit from 0 to 9. Initially, all panels show 0.

Every second, the digit shown by each panel increases by 1. In other words, at the end of every second, a panel that showed 9 would now show 0, a panel that showed 0 would now show 1, a panel that showed 1 would now show 2, and so on.

When a panel is paused, the digit displayed on the panel does not change in the subsequent seconds.

You must pause exactly one of these panels, at any second you wish. Then, the panels adjacent to it get paused one second later, the panels adjacent to those get paused 2 seconds later, and so on. In other words, if you pause panel x, panel y (for all valid y) would be paused exactly |x−y| seconds later.

For example, suppose there are 4 panels, and the 3-rd panel is paused when the digit 9 is on it.

  * The panel 1 pauses 2 seconds later, so it has the digit 1; 
  * the panel 2 pauses 1 second later, so it has the digit 0; 
  * the panel 4 pauses 1 second later, so it has the digit 0. 



The resulting 4-digit number is 1090. Note that this example is not optimal for n = 4.

Once all panels have been paused, you write the digits displayed on them from left to right, to form an n digit number (it can consist of leading zeros). What is the largest possible number you can get? Initially, all panels show 0.

Input

The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Each test case consists of a single line containing a single integer n (1 ≤ n ≤ 2⋅10^5).

It is guaranteed that the sum of n over all test cases does not exceed 2⋅10^5.

Output

For each test case, print the largest number you can achieve, if you pause one panel optimally.

Example

Input


2
1
2


Output


9
98

Note

In the first test case, it is optimal to pause the first panel when the number 9 is displayed on it.

In the second test case, it is optimal to pause the second panel when the number 8 is displayed on it.

Solution (in CPP):

Solution:
/***"In the name of Allah(swt), the most gracious, most merciful. Allah(swt) blesses with knowledge whom he wants."***/
/*** "ALHAMDULILLAH for Everything..." ***/
/*
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
*/

#include<bits/stdc++.h>
using namespace std;
#define  fastt       ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define  ll         long long
#define  debug      printf("Come here!")
#define  inf        1000000000000000000
#define  pb         push_back
#define  mp         make_pair
#define  sz(a)      a.size()
#define  mem(a,b)   memset(a,b,sizeof(a))
#define  clr(x)     x.clear()
#define  ff         first
#define  ss         second
#define  case(x)    cout << "Case " << x <<": "

#define  md         1000000007
#define  pi         acos(-1.0)
#define  cos(a)     cos(a*pi/180)
#define  sin(a)     sin(a*pi/180)
#define  tan(a)     tan(a*pi/180)
#define  cosi(a)    acos(a)/(pi/180)
#define  sini(a)    asin(a)/(pi/180)
#define  tani(a)    atan(a)/(pi/180)

#define  vl         vector <ll>
#define  vs         vector <string>
#define  vc         vector <char>
#define  vpll       vector < pair <ll, ll> >
#define  vpsl       vector < pair <string, ll> >
#define  vpls       vector < pair <ll, string> >
#define  pql        priority_queue<ll>
#define  minpql     priority_queue<ll,vector<ll>,greater<ll> >

#define  setbits(x)     __builtin_popcountll(x)
#define  zerobits(x)    __builtin_ctzll(x)

#define toll(a) atoll(a.c_str())
string tostr(ll a) {stringstream rr;rr<<a;return rr.str();}
ll pow(ll c,ll d){return d==0?:c*pow(c,d-1);}
ll gcd(ll a,ll b) {return b==0? a:gcd(b,a%b);}
ll lcm(ll a,ll b) {return ((a*b)/gcd(a,b));}
int dx[]={-1, 0, 0, 1};
int dy[]={0, -1, 1, 0};
int dx1[]={-1, -1, -1, 0, 0, 0, 1, 1, 1};
int dy1[]={-1, 0, 1, -1, 0, 1, -1, 0, 1};

/* mpp.max_load_factor(0.25); mpp.reserve(1024); */
/* cout << fixed << setprecision(12);*/

int main()
{
    fastt;
    ll t, z=0;
    cin>>t;
    while(t--){
        ll n, m, k, i, j, x, y, p=0, s=0, h, f=0;
        cin>>n;
        if(n==1)cout<<9<<endl;
        else if(n==2)cout<<98<<endl;
        else{
            cout<<98;
            n-=2;
            p=9;
            while(n--){
                cout<<p;
                p++;
                p%=10;
            }
            cout<<endl;
        }
    }
    return 0;
}