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 PYTHON3):

Solution:
import sys
from functools import reduce
from collections import Counter
import time
import datetime
import math 

# def time_t():
#     print("Current date and time: " , datetime.datetime.now())
#     print("Current year: ", datetime.date.today().strftime("%Y"))
#     print("Month of year: ", datetime.date.today().strftime("%B"))
#     print("Week number of the year: ", datetime.date.today().strftime("%W"))
#     print("Weekday of the week: ", datetime.date.today().strftime("%w"))
#     print("Day of year: ", datetime.date.today().strftime("%j"))
#     print("Day of the month : ", datetime.date.today().strftime("%d"))
#     print("Day of week: ", datetime.date.today().strftime("%A"))

def ip(): return int(sys.stdin.readline())

# def sip(): return sys.stdin.readline()
def sip() : return input()

def mip(): return map(int,sys.stdin.readline().split())

def mips(): return map(str,sys.stdin.readline().split())

def lip(): return list(map(int,sys.stdin.readline().split()))

def matip(n,m):
    lst=[]
    for i in range(n):
        arr = lip()
        lst.insert(i,arr)
    return lst

def factors(n): # find the factors of a number
    return list(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))

def minJumps(arr, n): #to reach from 0 to n-1 in the array in minimum steps
    jumps = [0 for i in range(n)]
    if (n == 0) or (arr[0] == 0):
        return float('inf')
    jumps[0] = 0
    for i in range(1, n):
        jumps[i] = float('inf')
        for j in range(i):
            if (i <= j + arr[j]) and (jumps[j] != float('inf')):
                jumps[i] = min(jumps[i], jumps[j] + 1)
                break
    return jumps[n-1]

def dic(arr): # converting list into dict of count
    return Counter(arr)

def check_prime(n):
    if n<2:
        return False
    for i in range(2,int(n**(0.5))+1,2):
        if n%i==0:
            return False
    return True 

# --------------------------------------------------------- #
# sys.stdin = open('input.txt','r')
# sys.stdout = open('output.txt','w')
# --------------------------------------------------------- #

t = ip()
while t:
    t-=1
    n = ip()
    print(9,end='')
    val = 8
    for i in range(n-1):
        print(val,end='')
        val+=1
        if val==10:
            val = 0
    print()
# print(time.process_time())