Good number(3进制)

https://codeforces.com/problemset/problem/1249/C2

C2. Good Numbers (hard version)

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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

The positive integer is called good if it can be represented as a sum of distinct powers of 33 (i.e. no duplicates of powers of 33 are allowed).

For example:

• 3030 is a good number: 30=33+3130=33+31,
• 11 is a good number: 1=301=30,
• 1212 is a good number: 12=32+3112=32+31,
• but 22 is not a good number: you can't represent it as a sum of distinct powers of 33 (2=30+302=30+30),
• 1919 is not a good number: you can't represent it as a sum of distinct powers of 33 (for example, the representations 19=32+32+30=32+31+31+31+3019=32+32+30=32+31+31+31+30 are invalid),
• 2020 is also not a good number: you can't represent it as a sum of distinct powers of 33 (for example, the representation 20=32+32+30+3020=32+32+30+30 is invalid).

Note, that there exist other representations of 1919 and 2020 as sums of powers of 33 but none of them consists of distinct powers of 33.

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

You have to answer qq independent queries.

Input

The first line of the input contains one integer qq (1≤q≤5001≤q≤500) — the number of queries. Then qqqueries follow.

The only line of the query contains one integer nn (1≤n≤10181≤n≤1018).

Output

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

Example

input

Copy

8
1
2
6
13
14
3620
10000
1000000000000000000

output

Copy

1
3
9
13
27
6561
19683
1350851717672992089

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdio.h>
#include <queue>
#include <stack>;
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 998244353
#define PI acos(-1)
using namespace std;
typedef long long ll ;
ll a[] ;

int main()
{
    int t ;
    scanf("%d" , &t);
    while(t--)
    {
        ll n ;
        scanf("%lld" , &n);
        memset(a ,  , sizeof(a));
        int cnt =  ;
        ll tmp = n ;
        while(tmp)
        {
            a[cnt++] = tmp %  ;
            tmp /=  ;

        }
        int pos = - ;
        for(int i =  ; i < cnt ; i++)
        {
            if(a[i] >= )
            {
                pos = i ;
                a[i] =  ;
                a[i+]++;
            }
        }
        if(pos != -)
        {
            for(int i =  ; i < pos ; i++)//确保最小
            {
                a[i] =  ;
            }
        }

        ll ans =  ;
        ll res =  ;
        for(int i =  ; i <= cnt ; i++)//注意取等
        {
            ans += a[i] * res;
            res *=  ;

        }
        cout << ans << endl ;

    }

    return  ;
}