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.
#include <bits/stdc++.h>
using namespace std;
bool is_prime(int p) {
  if (p == 1) {
    return false;
  }
  if (p == 2) {
    return true;
  }
  for (int i = 2; i * i <= p; ++i) {
    if (p % i == 0) {
      return false;
    }
  }
  return true;
}
const string s = "FastestFinger";
const string f = "Ashishgup";
void solve() {
  int n;
  cin >> n;
  if (n == 1) {
    cout << s << endl;
    return;
  }
  if (n == 2) {
    cout << f << endl;
    return;
  }
  if (n % 2 == 1) {
    cout << f << endl;
    return;
  }
  int tot2 = 1;
  while (n % 2 == 0) {
    n /= 2;
    tot2 *= 2;
  }
  if (n == 1) {
    cout << s << endl;
    return;
  }
  if (tot2 == 2 && is_prime(n)) {
    cout << s << endl;
    return;
  }
  cout << f << endl;
}
int main() {
  int t;
  cin >> t;
  for (int tcase = 0; tcase < t; ++tcase) {
    solve();
  }
  return 0;
}
