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 math import log2, ceil, floor, sqrt


def ans(n):
    if n == 1:
        return 'FastestFinger'
    if n % 2 == 1:
        return 'Ashishgup'
    if n == 2:
        return 'Ashishgup'
    #n is even
    # if n is a power of 2
    if ceil(log2(n)) == floor(log2(n)):
        return 'FastestFinger'
    if n % 4 == 0:
        return 'Ashishgup'
    # n is of the form 2*p
    p = int(n/2)
    for i in range(3, int(sqrt(p))+1, 2):
        if p % i == 0:
            return 'Ashishgup'
    return 'FastestFinger'


t = int(input())
L = []
for i in range(t):
    L.append(ans(int(input())))

for l in L:
    print(l)
