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;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tests = 1;
  cin >> tests;
  while (tests--) {
    long long int n;
    cin >> n;
    if (n % 2) {
      if (n == 1)
        cout << "FastestFinger" << endl;
      else
        cout << "Ashishgup" << endl;
      continue;
    }
    if (n == 2) {
      cout << "Ashishgup" << endl;
      continue;
    }
    long long int k = n / 2;
    long long int m = n;
    bool f = 0, p = 0;
    for (long long int i = 2; i * i <= n; i++) {
      while (n % i == 0) {
        if (i % 2) {
          f = 1;
          long long int l = (m / i) % 2;
          if (l == 0 && m / i > 2) p = 1;
        }
        n /= i;
      }
    }
    if (n != 1 && n % 2) {
      f = 1;
      long long int l = (m / n) % 2;
      if (l == 0 && m / n > 2) p = 1;
    }
    if (p)
      cout << "Ashishgup" << endl;
    else
      cout << "FastestFinger" << endl;
  }
  return 0;
}
