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:
#include <bits/stdc++.h>
#define int long long
#define INT_MAX INT64_MAX
#define INT_MIN INT64_MIN
#define PI 3.14159265358979323846
#define CLK clock_t clk = clock();//Start of main
#define OCLK cerr << "Time (in ms): " << (double)(clock() - clk) * 1000.0 / CLOCKS_PER_SEC << '\n';//End of main
#define MOD 1000000007
/* DEFINE cout */
#define pln cout << "===============================\n"
#define on cout<<"\n"
#define os cout<<" "
#define o2(a,b) cout<<a<<" "<<b
#define o(a) cout << a
#define bitcount __builtin_popcountll
/* DEFINE cout */
#define endl "\n"
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);//fastio
using namespace std;
struct hash_pair { // use pair as key in unordered_map<pair<int,int>,int,hash_pair>;
    template <class T1, class T2>
    size_t operator()(const pair<T1, T2>& p) const
    {
        auto hash1 = hash<T1>{}(p.first);
        auto hash2 = hash<T2>{}(p.second);
        return hash1 ^ hash2;
    }
};
bool prime[10000001];
int P=MOD-2;
int factorial[10000001]={0};
int powerFunction(int x,int y){
    int res = 1;
    int p = MOD;
    x = x % p;
    while (y > 0){
        if (y & 1)
            res = (res*x) % p;
        y = y>>1;
        x = (x*x) % p;
    }
    return res;
}
void sieveFunction(int maxLimit){
    memset(prime, true, sizeof(prime));
    prime[0] = prime[1]=false;
    for(int i=2; maxLimit >= i ; i++){
        if(prime[i]){
            for(int j=2*i ; maxLimit >= j ; j+=i)
            prime[j]=false;
        }
    }
}
void factorialFunction(int maxLimit){
    factorial[0]=1;
    for(int i=1 ; i <= maxLimit ; i++)
    factorial[i]=(factorial[i-1]*i)%MOD;
    return;
}
int gcd(int a, int b){
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
signed main(){
    int t;
    cin >> t;
    while(t--){
        int n;
        cin >> n;
        string str = "";
        str.push_back('9');
        int num = 8;
        for(int i=1; n>i; i++){
            str.push_back(num + '0');
            num++;
            num %= 10;
        }
        cout << str << endl;
        
    }
}