FACVSPOW - Factorial vs Power

Consider two integer sequences f(n) = n! and g(n) = an, where n is a positive integer. For any integer a > 1 the second sequence is greater than the first for a finite number of values. But starting from some integer k, f(n) is greater than g(n) for all n >= k. You are to find the least positive value of n for which f(n) > g(n), for a given positive integer a > 1.

Input

The first line of the input contains number t – the amount of tests. Then t test descriptions follow. Each test consist of a single number a.

Constraints

1 ≤ t ≤ 100000
2 ≤ a ≤ 106

Output

For each test print the least positive value of n for which f(n) > g(n).

Example

Input:
3
2
3
4

Output:
4
7
9

Added by:Spooky
Date:2009-11-01
Time limit:1s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64 NODEJS OBJC PERL6 SQLITE VB.NET
Resource:Advancement Autumn 2009, http://sevolymp.uuuq.com/

hide comments
2014-04-30 07:21:01 sarelfeniel
Good problem. No idea how the fastest people are doing it. Must be secrets of mathematics I do not know.

Some good test cases from the forum that helped me

--edit(francky)--> Please do not add test cases, those in description should be enough.

@Francky I guess... honestly it takes < 3 seconds to go to the forum to find additional cases.

Last edit: 2014-05-04 10:11:40
2013-07-01 09:42:20 Hasil Sharma
newton raphson or something else to be used by any chance ?
2013-04-11 08:10:52 raunakrocks
esy math...no its programming..AC!!
2013-02-17 07:43:29 Inspiron
nice prob....
2012-07-20 14:35:23 devu
Sab Funde Baji Hai .. Think Programming ,not mathematics ,mathematics part is simple....
2011-03-10 13:51:08 sandeep pandey
avoid stirling appx.(due to precision issue).
think mathematics:-))
2011-02-01 07:51:45 bashrc is back
I am getting TLE.....Am i missing something?
2010-11-03 11:45:04 jacob
can anyone with an AC give me some test cases for very large numbers ~10^5 or ~10^6
i am getting a wrong answer using stirling approximation
2010-08-29 09:11:52 Curiosa
dont calculate n! and a^n. just use some mathematics
2010-06-25 00:17:59 :(){ :|: & };:

It is possible to get an Accepted verdict by using Stirling approximation :)

Last edit: 2010-06-25 00:19:42
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