Ashishgup and FastestFinger play a game. 

They start with a number n and play in turns. In each turn, a player can make any one of the following moves:

  * Divide n by any of its odd divisors greater than 1. 
  * Subtract 1 from n if n is greater than 1. 



Divisors of a number include the number itself.

The player who is unable to make a move loses the game.

Ashishgup moves first. Determine the winner of the game if both of them play optimally.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.

The only line of each test case contains a single integer — n (1 ≤ n ≤ 10^9).

Output

For each test case, print "Ashishgup" if he wins, and "FastestFinger" otherwise (without quotes).

Example

Input


7
1
2
3
4
5
6
12


Output


FastestFinger
Ashishgup
Ashishgup
FastestFinger
Ashishgup
FastestFinger
Ashishgup

Note

In the first test case, n = 1, Ashishgup cannot make a move. He loses.

In the second test case, n = 2, Ashishgup subtracts 1 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.

In the third test case, n = 3, Ashishgup divides by 3 on the first move. Now n = 1, FastestFinger cannot make a move, so he loses.

In the last test case, n = 12, Ashishgup divides it by 3. Now n = 4, FastestFinger is forced to subtract 1, and Ashishgup gets 3, so he wins by dividing it by 3.
def isPrime(n):
	i = 2
	while i*i <= n:
		if n%i == 0:
			return False
		i += 1
	return True

def solve(n):

	if n == 1:
		win = 2
	elif n == 2:
		win = 1
	elif n%2 != 0:
		# player 1 will divide by odd = n and turn n = 1
		win = 1
	elif n&(n-1) == 0:
		# if n is power of 2 then only option left is -1
		# after this n = odd and player 2 win
		win = 2
	else:
		# now player 1 will lose only if n = 2 * single_prime
		# in such case player 1 is lose in both cases (-1 or / by prime)

		# we are sure n in even here if only 2^1 is present then n%4 != 0
		# and n//2 is prime

		# if 2 power is atlest 2 or more than player 1 can win by giving player 2
		# num = 2^x and divide off all odd prime factors so 2 had only option -1
		# which make n = odd and 1 win

		# 1 can also win if num = 2 * odd1 * odd2 * .. * oddn
		# here 1 will offer num = 2 * odd1 to player 2 he have only option is to
		# divide and n = 2, player 1 win 

		# thus overall a player here lose only if he has n = 2 * single_prime
		if n%4 != 0 and isPrime( n//2 ):
			win = 2
		else:
			win = 1

	if win == 2:	
		return 'FastestFinger'
	else:
		return 'Ashishgup'  


for _ in range(int(input())):
	n = int(input())
	print( solve(n) )