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 floor

def is_odd_prime(n):
    if n % 3 == 0:
        n //= 3
        if n == 1:
            return True
        else:
            return False

    f = 6
    while f < floor(n ** 0.5) + 2:
        for i in (f-1, f+1):
            if n % i == 0:
                n //= i
                if n == 1:
                    return True
                else:
                    return False
        f += 1

    return True

def f2(n):
    power = 0
    while n % 2 == 0:
        power += 1
        n //= 2

    return power, n

def solve(n):
    F, A = "FastestFinger", "Ashishgup"
    if n == 1:
        return F
    elif (n % 2 == 1) or n == 2:
        return A
    else:
        pow_2, odd_n = f2(n)
        if odd_n == 1:
            return F
        elif pow_2 == 0:
            return A
        elif pow_2 == 1:
            if is_odd_prime(odd_n):
                return F

        return A       

t = int(input())

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