The only difference between easy and hard versions is the maximum value of n.

You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n.

The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed).

For example:

  * 30 is a good number: 30 = 3^3 + 3^1, 
  * 1 is a good number: 1 = 3^0, 
  * 12 is a good number: 12 = 3^2 + 3^1, 
  * but 2 is not a good number: you can't represent it as a sum of distinct powers of 3 (2 = 3^0 + 3^0), 
  * 19 is not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representations 19 = 3^2 + 3^2 + 3^0 = 3^2 + 3^1 + 3^1 + 3^1 + 3^0 are invalid), 
  * 20 is also not a good number: you can't represent it as a sum of distinct powers of 3 (for example, the representation 20 = 3^2 + 3^2 + 3^0 + 3^0 is invalid). 



Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3.

For the given positive integer n find such smallest m (n ≤ m) that m is a good number.

You have to answer q independent queries.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 500) — the number of queries. Then q queries follow.

The only line of the query contains one integer n (1 ≤ n ≤ 10^4).

Output

For each query, print such smallest integer m (where n ≤ m) that m is a good number.

Example

Input


7
1
2
6
13
14
3620
10000


Output


1
3
9
13
27
6561
19683
"""
    Satwik_Tiwari ;) .
    16 june , 2020  - Tuesday
"""

#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function

from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase


import bisect
import heapq
from math import *
from collections import deque
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned

#==============================================================================================

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

#===============================================================================================
#some shortcuts

mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for p in range(t):
        solve()
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(a,b):
    ans = 1
    while(b>0):
        if(b%2==1):
            ans*=a
        a*=a
        b//=2
    return ans
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) : # Check Prime Number or not
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True

#===============================================================================================
# code here ;))

def solve():
    n = int(inp())
    x = 40
    num = 0
    temp = n
    helpp = -1
    new = -1
    f = False
    while(temp > 0 and x >= 0):
        # print(temp,3**x)
        if(temp < 3**x):
            if(f):
                new = x
            x-=1
        else:
            if(helpp == -1):
                helpp = x
            f = True
            num += 3**x
            temp -= 3**x
            x-=1

    if(temp == 0):
        print(num)
    else:
        if(x >= 0):
            print(num+1)
        else:
            if(new == -1):
                print(3**(helpp+1))
            else:
                ans = num + 3**new
                x = 0
                while(x < new):
                    ans -= 3**x
                    x+=1
                print(ans)




# testcase(1)
testcase(int(inp()))













