FAST2  Fast Sum of two to an exponent
There is people that really like to do silly thinks, one of them is sum the numbers from 2^0 to 2^n, task is actually really simple, just do a ultra fast sum of term 2^0 to 2^n
Input
the first line starts with a number, T, wich is the number of test cases, T lines will follow
each line contains a number "n" that is the nth term of the sum from 2^0 to 2^n
0<=n<=500
Output
Output the sum from 2^0 to 2^n MODULO 1298074214633706835075030044377087
Example
Input: 3
0
1
2
Output: 1
3
7
Extra: TLE is equal to 0.15s
hide comments
aj_254:
20190423 21:06:55
one line code in python just get sum of series that 2**0 +2**1+...2**n=2**(n+1)1 remember this formula 

charlles:
20190410 17:52:22
bitwise operations turn it easy. ;) 

imkiller:
20180601 15:15:14
Use pefixSum and precompute it so that u can retrive in O(1)


dragon_fury:
20151103 16:10:58
python rocks...... :D 

Thotsaphon Thanatipanonda:
20151021 09:30:43
Finally, Java can pass this problem!! :D 

dwij28:
20151001 16:29:10
Used sys library in python for input, input() function results in TLE, also modular exponentiation gives TLE .. A hint here : think of the fastest way to find 2 ** n, there is a very fast method to do this, and you do not require any exponent functions for this .. :) 

gullu_mishra:
20150918 18:28:04
Lovely python ... u rock baby ;)


Parul Yadav:
20150907 13:23:36
my first q in python 

Vladimir:
20150805 17:42:06
Edit: nevermind Last edit: 20150805 17:45:17 

(Tjandra Satria Gunawan)(æ›¾æ¯…æ˜†):
20120504 02:53:09
No need to use BigInt ;)

Added by:  Rocker3011 
Date:  20120414 
Time limit:  0.100s 
Source limit:  500B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: ASM64 
Resource:  Own problem 