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;
void c_p_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void SieveOfEratosthenes(long long n, vector<long long> a, long long count) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long p = 2; p <= n; p++)
    if (prime[p]) {
      a.push_back(p);
      count++;
    }
}
int32_t main() {
  c_p_c();
  long long T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    long long m = n;
    long long count = 0;
    while (m > 0) {
      count++;
      m /= 2;
    }
    if (n == 1) {
      cout << "FastestFinger" << endl;
      continue;
    }
    if (n == 2) {
      cout << "Ashishgup" << endl;
      continue;
    }
    if (pow(2, count - 1) == n) {
      cout << "FastestFinger" << endl;
      continue;
    }
    if (n % 2 == 0 && (n / 2) % 2 == 1) {
      long long k = n / 2;
      long long prime = 1;
      for (long long i = 3; i <= pow(n, 0.5); i += 2) {
        if (k % i == 0) {
          prime = 0;
          break;
        }
      }
      if (prime == 1) {
        cout << "FastestFinger" << endl;
        continue;
      } else {
        cout << "Ashishgup" << endl;
        continue;
      }
    } else {
      cout << "Ashishgup" << endl;
    }
  }
  return 0;
}
