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:
from __future__ import division, print_function
from itertools import permutations, combinations
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**7)
# sys.stdin = open('inpy.txt', 'r')
# sys.stdout = open('outpy.txt', 'w')
from sys import stdin, stdout
import bisect  # c++ upperbound
import math
import heapq
i_m = 9223372036854775807


def inin():
    return int(input())


def stin():
    return input()


def spin():
    return map(int, stin().split())


def lin():  # takes array as input
    return list(map(int, stin().split()))


def matrix(n):
    #matrix input
    return [list(map(int, input().split()))for i in range(n)]

################################################


def string2intlist(s):
    return list(map(int, s))


def calculate_sum(a, N):  # sum of a to N
    # Number of multiples
    m = N / a
    # sum of first m natural numbers
    sum = m * (m + 1) / 2
    # sum of multiples
    ans = a * sum
    return ans


def series(N):
    return (N*(N+1))//2


def count2Dmatrix(i, list):
    return sum(c.count(i) for c in list)


def modinv(n, p):
    return pow(n, p - 2, p)


def nCr(n, r):
    i = 1
    while i < r:
        n *= (n - i)
        i += 1
    return n // math.factorial(r)

def LCM(x, y):
    return (x * y) // math.gcd(x, y)


def Divisors(n):
    l = []
    for i in range(1, int(math.sqrt(n) + 1)):
        if (n % i == 0):
            if (n // i == i):
                l.append(i)
            else:
                l.append(i)
                l.append(n//i)
    return l


def isprime(n):
    if n == 2 or n == 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False
    for i in range(5, int(math.sqrt(n))+1, 6):
        if n % i == 0 or n % (i+2) == 0:
            return False
    return True


prime = []


def SieveOfEratosthenes(n):
    global prime
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    f = []
    for p in range(2, n):
        if prime[p]:
            f.append(p)
    return f


q = []


def dfs(n, d, v, c):
    global q
    v[n] = 1
    x = d[n]
    q.append(n)
    j = c
    for i in x:
        if i not in v:
            f = dfs(i, d, v, c+1)
            j = max(j, f)
            # print(f)
    return j
# d = {}


def knapSack(W, wt, val, n):
    K = [[0 for x in range(W + 1)] for x in range(n + 1)]

    # Build table K[][] in bottom up manner
    for i in range(n + 1):
        for w in range(W + 1):
            if i == 0 or w == 0:
                K[i][w] = 0
            elif wt[i-1] <= w:
                K[i][w] = max(val[i-1] + K[i-1][w-wt[i-1]],  K[i-1][w])
            else:
                K[i][w] = K[i-1][w]

    return K[n][W]


def modularExponentiation(x, n):
    M = 10**9+7
    if(n == 0):
        return 1
    elif (n % 2 == 0):  # n is even
        return modularExponentiation((x*x) % M, n//2)
    else:  # n is odd
        return (x * modularExponentiation((x * x) % M, (n - 1) // 2)) % M


def modInverse(a, m):
    m0 = m
    y = 0
    x = 1

    if (m == 1):
        return 0

    while (a > 1):

        # q is quotient
        q = a // m

        t = m

        # m is remainder now, process
        # same as Euclid's algo
        m = a % m
        a = t
        t = y

        # Update x and y
        y = x - q * y
        x = t

    # Make x positive
    if (x < 0):
        x = x + m0

    return x


"""*******************************************************"""
for _ in range(inin()):
    n = inin()

    if n == 1:
        print(9)
    elif n == 2:
        print(98)
    elif n == 3:
        print(989)
    else:
        print(989, end="")
        j = 0
        for i in range(0, n - 3):
            if j == 10:
                j = 0
            print(j, end="")
            j += 1
        print()