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 global = 0;
void test_case(long long);
int main() {
  clock_t begin, end;
  begin = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  int yes = 1;
  if (yes)
    cin >> t;
  else
    t = 1;
  for (long long(T) = 0; (T) < t; (T)++) test_case(T);
  end = clock();
  return 0;
}
void test_case(long long T) {
  long long n;
  cin >> n;
  string b = "FastestFinger";
  string a = "Ashishgup";
  if (n == 1) {
    cout << b << endl;
    return;
  }
  if (n == 2) {
    cout << a << endl;
    return;
  }
  if (n % 2) {
    cout << a << endl;
    return;
  }
  long long p = 0;
  while (n % 2 == 0 && n > 1) {
    n /= 2;
    p++;
  }
  if (n == 1) {
    if (p == 1)
      cout << a << endl;
    else
      cout << b << endl;
  } else {
    long long lim = sqrt(n) + 1;
    for (long long j = 2; j < lim; j++) {
      if (n % j == 0) {
        cout << a << endl;
        return;
      }
    }
    if (p == 1)
      cout << b << endl;
    else
      cout << a << endl;
    return;
  }
}
