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
import sys
import bisect as bi
import math
from collections import defaultdict as dd
import heapq
import itertools
input=sys.stdin.readline
from random import randint
##import numpy as np
##sys.setrecursionlimit(10**7)
mo=10**9+7
def cin():
    return map(int,sin().split())
def ain():            
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
def d3(n): 
    Num =[] 
    i = 0; 
    while (n > 0):  
        Num+= [n % 3]; 
        n = int(n / 3); 
        i += 1;
    return Num[::-1]
dp=[0]*40
dp[0]=1
for i in range(1,40):
    dp[i]=dp[i-1]*3    
for _ in range(inin()):
    n=inin()
    s=d3(n)
    n1=len(s)
##    print(s)
    if(2 in s and n not in dp):
        c2=0;p2=0;pz=n1;f1=f=1
        for i in range(n1):
            if(s[i]==2):
                if(f):
                    p2=i
                    f=0
                c2+=1
            elif(s[i]==0):
                if(c2==0):
                    pz=i
                    
##        print(p2,pz)
        if(pz>p2):print(dp[n1])
        else:
            s=s[0:pz]+[1]+[0]*(n1-pz+1)
##            print(s)
            ans=0
            for i in range(n1-1,-1,-1):
                ans+=s[n1-i-1]*dp[i]
            print(ans)        
    else:
        print(n)
    
##    if((n*(n+1)//2)%2):print("NO")
##    else:
##        print("YES")
##        print(n//2)
##        print(l[
        
        
        
    
        
    
        
      

        
        
    
    
    
 
    
    
            
    
        
    
    
    

                    
                    
                    
                    
                    
        
    
    
    
    
            
        

##def msb(n):n|=n>>1;n|=n>>2;n|=n>>4;n|=n>>8;n|=n>>16;n|=n>>32;n|=n>>64;return n-(n>>1) #2 ki power
##def pref(a,n,f):             
##    pre=[0]*n
##    if(f==0):         ##from beginning
##        pre[0]=a[0]
##        for i in range(1,n):
##            pre[i]=a[i]+pre[i-1]
##    else:              ##from end
##        pre[-1]=a[-1]
##        for i in range(n-2,-1,-1):
##            pre[i]=pre[i+1]+a[i]
##    return pre
##maxint=10**24 
##def kadane(a,size): 
##    max_so_far = -maxint - 1
##    max_ending_here = 0
##       
##    for i in range(0, size): 
##        max_ending_here = max_ending_here + a[i] 
##        if (max_so_far < max_ending_here): 
##            max_so_far = max_ending_here 
##  
##        if max_ending_here < 0: 
##            max_ending_here = 0   
##    return max_so_far
##def power(x, y):
##    if(y == 0):return 1
##    temp = power(x, int(y / 2))%mo
##    if (y % 2 == 0):return (temp * temp)%mo 
##    else:
##        if(y > 0):return (x * temp * temp)%mo 
##        else:return ((temp * temp)//x )%mo

##for _ in range(1):
##    n=inin()
##    l=ain()
##    d=dd(list)
##    dp=dd(set)
##    for k in range(n):
##        i=l[k]
##        dp[i].add(k)
##        b=bin(i)[2:]
##        c=0
##        for j in b[::-1]:
##            if j=='1':
##                d[c]+=[[i,k]]
##            c+=1
##    key=sorted(list(d.keys()),reverse=True)
##    ans=0
##    for i in key:
##        va=d[i]
##        va.sort(reverse=True)
##        p=len(va)
##        if(p!=0):
##            c=0;tans=0
##            for j,po in va:
##                if(po in dp[j]):
##                    c+=1
##                    f1,f2=j,po
##                    tans+=j
##                    dp[j].remove(po)
##            if(c%2==0 and c!=0):
##                c-=1
##                
##            ans+=tans
##            print(ans,i,c,va)        
##    print(ans)
##        
##        
##    
    
    
    
    
        
        
    
    
    
    


    




##def msb(n):n|=n>>1;n|=n>>2;n|=n>>4;n|=n>>8;n|=n>>16;n|=n>>32;n|=n>>64;return n-(n>>1) #2 ki power
##def pref(a,n,f):             
##    pre=[0]*n
##    if(f==0):         ##from beginning
##        pre[0]=a[0]
##        for i in range(1,n):
##            pre[i]=a[i]+pre[i-1]
##    else:              ##from end
##        pre[-1]=a[-1]
##        for i in range(n-2,-1,-1):
##            pre[i]=pre[i+1]+a[i]
##    return pre
##maxint=10**24 
##def kadane(a,size): 
##    max_so_far = -maxint - 1
##    max_ending_here = 0
##       
##    for i in range(0, size): 
##        max_ending_here = max_ending_here + a[i] 
##        if (max_so_far < max_ending_here): 
##            max_so_far = max_ending_here 
##  
##        if max_ending_here < 0: 
##            max_ending_here = 0   
##    return max_so_far
##def power(x, y):
##    if(y == 0):return 1
##    temp = power(x, int(y / 2))%mo
##    if (y % 2 == 0):return (temp * temp)%mo 
##    else:
##        if(y > 0):return (x * temp * temp)%mo 
##        else:return ((temp * temp)//x )%mo

            
        
        
        
            
    
        
        
    
    
    

 
    
    
    
    
        
        
    
    
    
    


    




##def msb(n):n|=n>>1;n|=n>>2;n|=n>>4;n|=n>>8;n|=n>>16;n|=n>>32;n|=n>>64;return n-(n>>1) #2 ki power
##def pref(a,n,f):             
##    pre=[0]*n
##    if(f==0):         ##from beginning
##        pre[0]=a[0]
##        for i in range(1,n):
##            pre[i]=a[i]+pre[i-1]
##    else:              ##from end
##        pre[-1]=a[-1]
##        for i in range(n-2,-1,-1):
##            pre[i]=pre[i+1]+a[i]
##    return pre
##maxint=10**24 
##def kadane(a,size): 
##    max_so_far = -maxint - 1
##    max_ending_here = 0
##       
##    for i in range(0, size): 
##        max_ending_here = max_ending_here + a[i] 
##        if (max_so_far < max_ending_here): 
##            max_so_far = max_ending_here 
##  
##        if max_ending_here < 0: 
##            max_ending_here = 0   
##    return max_so_far
##def power(x, y):
##    if(y == 0):return 1
##    temp = power(x, int(y / 2))%mo
##    if (y % 2 == 0):return (temp * temp)%mo 
##    else:
##        if(y > 0):return (x * temp * temp)%mo 
##        else:return ((temp * temp)//x )%mo
