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.
import math

WIN = "Ashishgup"
LOSS = "FastestFinger"

def sumpfac (n):
    assert n % 2 != 0
    sum = 0
    cont = True
    while (cont):
        cont = False
        for i in range (3, int (math.sqrt (n))+1, 2):
            if n % i == 0:
                n //= i
                sum += 1
                cont = True
                break
    if n > 1:
        sum += 1
    return sum

def solve (n):
    if n == 1:
        #print ('|1')
        return LOSS
    if n % 2 == 1:
        #print ('|odd')
        return WIN
    if (n == 2):
        #print ('|2')
        return WIN

    t = n
    pow2 = 0
    while t % 2 == 0:
        t //= 2
        pow2 += 1

    if pow2 == 1:
        if (sumpfac (t) >= 2):
            #print ('|2.p^x')
            return WIN
        else:
            #print ('|2.p')
            return LOSS

    if t == 1:
        #print ('|2^x')
        return LOSS

    #print ('|2^x.m')
    return WIN

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