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.
'''input
7
1
2
3
4
5
6
12


'''
from collections import defaultdict
import math
from functools import reduce
from heapq import heapify, heappush, heappop
import sys
from sys import stdin, stdout
from random import seed
from random import randint


mod = 10 ** 9 + 7


def mul_mod(a, b):
    return ((a % mod) * (b % mod)) % mod


def add_mod(a, b):
    return ((a % mod) + (b % mod)) % mod

def sub_mod(a, b):
	return((a % mod) - (b % mod) + mod) % mod

def modInverse(b):
    # g = math.gcd(b, m)
    # if (g != 1):
    #     # print("Inverse doesn't exist")
    #     return -1
    return pow(b, mod - 2, mod)

def div_mod(a, b):
    a = a % mod
    inv = modInverse(b)
    # if(inv == -1):
    #     print("Division not defined")
    # else:
    #     print("Result of Division is ",(inv*a) % m)
    return (inv * a) % mod


# This class represents an undirected graph using adjacency list representation
class Graph:

    def __init__(self, vertices):
        self.V = vertices + 1  # No. of vertices
        self.graph = defaultdict(set)  # default dictionary to store graph
        self.Time = 0
        self.P = 0
        self.Q = 0

    # function to add an edge to graph
    def addEdge(self, u, v):
        self.graph[u].add(v)
        self.graph[v].add(u)

    '''A recursive function that finds and prints bridges
    using DFS traversal
    u --> The vertex to be visited next
    visited[] --> keeps tract of visited vertices
    disc[] --> Stores discovery times of visited vertices
    parent[] --> Stores parent vertices in DFS tree
    count[] --> Stores count of vertices in DFS subtree rooted at the vertex'''

    def bridgeUtil(self, u):
        # Mark the current node as visited and print it
        # self.visited[u] = True
        self.count[u] += 1

        # Initialize discovery time and low value
        self.disc[u] = self.Time
        self.low[u] = self.Time
        self.Time += 1

        # Recur for all the vertices adjacent to this vertex
        for v in self.graph[u]:
            # If v is not visited yet, then make it a child of u
            # in DFS tree and recur for it
            if self.count[v] == 0:
                self.parent[v] = u
                self.bridgeUtil(v)
                self.count[u] += self.count[v]

                # Check if the subtree rooted with v has a connection to
                # one of the ancestors of u
                self.low[u] = min(self.low[u], self.low[v])

                ''' If the lowest vertex reachable from subtree
                under v is below u in DFS tree, then u-v is
                a bridge'''
                if self.low[v] > self.disc[u]:
                    self.Q += 1
                    if self.count[v] % 2 == 0:
                        self.P += 1
                        # print("%d %d %d" % (u, v, count[v]))

            # Update low value of u for parent function calls.
            elif v != self.parent[u]:
                self.low[u] = min(self.low[u], self.disc[v])

    # DFS based function to find all bridges. It uses recursive
    # function bridgeUtil()
    def bridge(self):

        # Mark all the vertices as not visited and Initialize parent and visited,
        # and ap(articulation point) arrays
        # self.visited = [False] * (self.V)
        self.disc = [float("Inf")] * (self.V)
        self.low = [float("Inf")] * (self.V)
        self.parent = [-1] * (self.V)
        self.count = [0] * (self.V)

        # Call the recursive helper function to find bridges
        # in DFS tree rooted with vertex 'i'
        for i in range(1, self.V):
            if self.count[i] == 0:
                self.bridgeUtil(i)
        # print(count)

def prime(n):
	i = 2
	while i * i <= n:
		if n % i == 0:
			return False
		i += 1
	return True



for _ in range(int(input())):
    n = int(input())
    # a = list(map(int, input().split()))
    # b = list(map(int, input().split()))
    # c, d = reversed(a), reversed(b)
    # e, f = zip(a, c), zip(b, d)
    # e, f = sorted(e), sorted(f)
    # pow_2 = [2]
    if n == 1:
    	print("FastestFinger")
    	continue
    if n == 2:
    	print("Ashishgup")
    	continue
    if n % 2 != 0:
    	print("Ashishgup")
    	continue
    # if n % 4 != 0:
    # 	print("FastestFinger")
    # else:
    # 	print("Ashishgup")
    if not(n & (n - 1)):
    	print("FastestFinger")
    else:
    	if n % 4 != 0:
    		n = n // 2
    		if prime(n):
    			print("FastestFinger")
    		else:
    			print("Ashishgup")
    	else:
    		print("Ashishgup")

