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.
# from collections import defaultdict
import math

rr = lambda: input()
rri = lambda: int(input())
rrm = lambda: list(map(int, input().split()))

INF=float('inf')

def solve(n):
	# odd -> lost since person will divide by self to get 1
	# even -> odd divisor (if we subtract 1, lose next turn)
	orig = n

	A = "Ashishgup"
	B = "FastestFinger"

	if n == 1:
		return B
	elif n == 2:
		return A
	elif n&1:
		return A

	# even number needs at least 1 odd divisor (not 1), else Ashishgup loses
	leftover = 1 # amount left after removing odd divisor
	while not n&1:
		n //= 2 # n is largest odd factor removed
		leftover *= 2

	# check if n has more than 1 odd factor, otherwise A can alawys win
	for i in range(3,int(math.sqrt(n)+1),2):
		if n%i==0:
			return A
	if n == 1: # 1 has no other odd factor
		# no other odd factors
		return B
	elif leftover == 2:
		return B
	else:
		return A

t = rri()
for _ in range(t):
	n=rri()
	print(solve(n))
