You are given an integer n. Find any string s of length n consisting only of English lowercase letters such that each non-empty substring of s occurs in s an odd number of times. If there are multiple such strings, output any. It can be shown that such string always exists under the given constraints.

A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Input

The first line contains a single integer t (1 ≤ t ≤ 500) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).

It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.

Output

For each test case, print a single line containing the string s. If there are multiple such strings, output any. It can be shown that such string always exists under the given constraints.

Example

Input


4
3
5
9
19


Output


abc
diane
bbcaabbba
youarethecutestuwuu

Note

In the first test case, each substring of "abc" occurs exactly once.

In the third test case, each substring of "bbcaabbba" occurs an odd number of times. In particular, "b" occurs 5 times, "a" and "bb" occur 3 times each, and each of the remaining substrings occurs exactly once.
#DaRk DeveLopeR
import sys


#taking input as string 
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
mod = 10**9+7; Mod = 998244353; INF = float('inf')
#______________________________________________________________________________________________________
import math
from bisect import *
from heapq import *
from collections import defaultdict as dd
from collections import OrderedDict as odict
from collections import Counter as cc
from collections import deque
from itertools import groupby
sys.setrecursionlimit(20*20*20*20+10) #this is must for dfs



def solve():

	n=takein()

	if n==1:
		print("a")
		return 


	ans="a"*(n//2)

	if n%2==0:
		ans+='b'
	else:
		ans+="bc"
	ans+="a"*((n//2)-1)

	print(ans)

	




def main():
    global tt
    if not ONLINE_JUDGE:
        sys.stdin = open("input.txt","r")
        sys.stdout = open("output.txt","w")
    t = 1
    t = takein()
    #t = 1
    for tt in range(1,t + 1):
        solve()
    if not ONLINE_JUDGE:
        print("Time Elapsed :",time.time() - start_time,"seconds")
        sys.stdout.close()

#---------------------- USER DEFINED INPUT FUNCTIONS ----------------------#
def takein():
    return (int(sys.stdin.readline().rstrip("\r\n")))


# input the string

def takesr():
    return (sys.stdin.readline().rstrip("\r\n"))


# input int array
def takeiar():
    return (list(map(int, sys.stdin.readline().rstrip("\r\n").split())))


# input string array
def takesar():
    return (list(map(str, sys.stdin.readline().rstrip("\r\n").split())))


# innut values for the diffrent variables
def takeivr():
    return (map(int, sys.stdin.readline().rstrip("\r\n").split()))


def takesvr():
    return (map(str, sys.stdin.readline().rstrip("\r\n").split()))

 
 
#------------------ USER DEFINED PROGRAMMING FUNCTIONS ------------------#



def ispalindrome(s):

	return s==s[::-1]
 
def invert(bit_s):


	  
	# convert binary string 
	# into integer
	temp = int(bit_s, 2)
	  
	# applying Ex-or operator
	# b/w 10 and 31
	inverse_s = temp ^ (2 ** (len(bit_s) + 1) - 1)
	  
	# convert the integer result 
	# into binary result and then 
	# slicing of the '0b1' 
	# binary indicator
	rslt = bin(inverse_s)[3 : ]

	return str(rslt) 	  

def counter(a):
    q = [0] * max(a)
    for i in range(len(a)):
        q[a[i] - 1] = q[a[i] - 1] + 1
    return(q)
 
def counter_elements(a):
    q = dict()
    for i in range(len(a)):
        if a[i] not in q:
            q[a[i]] = 0
        q[a[i]] = q[a[i]] + 1
    return(q)
 
def string_counter(a):
    q = [0] * 26
    for i in range(len(a)):
        q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
    return(q)
 
def factorial(n,m = 1000000007):
    q = 1
    for i in range(n):
        q = (q * (i + 1)) % m
    return(q)
 
def factors(n):
    q = []
    for i in range(1,int(n ** 0.5) + 1):
        if n % i == 0: q.append(i); q.append(n // i)
    return(list(sorted(list(set(q)))))
 
def prime_factors(n):
    q = []
    while n % 2 == 0: q.append(2); n = n // 2
    for i in range(3,int(n ** 0.5) + 1,2):
        while n % i == 0: q.append(i); n = n // i
    if n > 2: q.append(n)
    return(list(sorted(q)))
 
def transpose(a):
    n,m = len(a),len(a[0])
    b = [[0] * n for i in range(m)]
    for i in range(m): 
        for j in range(n): 
            b[i][j] = a[j][i]
    return(b)
 
def power_two(x):
    return (x and (not(x & (x - 1))))
 
def ceil(a, b):
    return -(-a // b)



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

 
def pref(li):
    pref_sum = [0]
    for i in li:
        pref_sum.append(pref_sum[-1]+i)
    return pref_sum
 
def kadane(x):  # maximum sum contiguous subarray
    sum_so_far = 0
    current_sum = 0
    for i in x:
        current_sum += i
        if current_sum < 0:
            current_sum = 0
        else:
            sum_so_far = max(sum_so_far, current_sum)
    return sum_so_far

def binary_search(li, val):
    # print(lb, ub, li)
    ans = -1
    lb = 0
    ub = len(li)-1
    while (lb <= ub):
        mid = (lb+ub) // 2
        # print('mid is',mid, li[mid])
        if li[mid] > val:
            ub = mid-1
        elif val > li[mid]:
            lb = mid+1
        else:
            ans = mid  # return index
            break
    return ans


 
def upper_bound(li, num):
    answer = -1
    start = 0
    end = len(li)-1
 
    while (start <= end):
        middle = (end+start) // 2
 
        if li[middle] <= num:
            answer = middle
            start = middle+1
 
        else:
            end = middle-1
    return answer  # max index where x is not greater than num


def lower_bound(li, num):
    answer = -1
    start = 0
    end = len(li)-1
 
    while (start <= end):
        middle = (end+start) // 2
        if li[middle] >= num:
            answer = middle
            end = middle-1
        else:
            start = middle+1
    return answer  # min index where x is not less than num
#-----------------------------------------------------------------------#
ONLINE_JUDGE = __debug__
if ONLINE_JUDGE:
    input = sys.stdin.readline
    
main()