#include <bits/stdc++.h>
using namespace std;
bool isprime(long long n) {
  long long i;
  if (n == 2 || n == 3) return 1;
  long long flag = 0;
  for (i = 2; i <= sqrt(n) + 1; i++) {
    if (n % i == 0) {
      flag = 1;
      break;
    }
  }
  if (flag)
    return 0;
  else
    return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    if (n == 1) {
      cout << "FastestFinger"
           << "\n";
      continue;
    }
    if (n % 2 == 1) {
      cout << "Ashishgup"
           << "\n";
      continue;
    }
    long long temp = n, i = 2, c = 0;
    long long c1 = 0;
    while (n % 2 == 0) {
      n = n / i;
      c++;
    }
    if (n == 1 && temp != 2) {
      cout << "FastestFinger"
           << "\n";
      continue;
    }
    if (n != 1 && isprime(n)) {
      c1++;
    } else
      c1 = 2;
    if (c == 1 && c1 == 1) {
      cout << "FastestFinger"
           << "\n";
    } else if (c == 1 && c1 > 1) {
      cout << "Ashishgup"
           << "\n";
    } else if (c == 1 && c1 == 0) {
      cout << "Ashishgup"
           << "\n";
    } else if (c1 == 0) {
      cout << "FastestFinger"
           << "\n";
    } else {
      cout << "Ashishgup"
           << "\n";
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
bool isprime(long long n) {
  long long i;
  if (n == 2 || n == 3) return 1;
  long long flag = 0;
  for (i = 2; i <= sqrt(n) + 1; i++) {
    if (n % i == 0) {
      flag = 1;
      break;
    }
  }
  if (flag)
    return 0;
  else
    return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    if (n == 1) {
      cout << "FastestFinger"
           << "\n";
      continue;
    }
    if (n % 2 == 1) {
      cout << "Ashishgup"
           << "\n";
      continue;
    }
    long long temp = n, i = 2, c = 0;
    long long c1 = 0;
    while (n % 2 == 0) {
      n = n / i;
      c++;
    }
    if (n == 1 && temp != 2) {
      cout << "FastestFinger"
           << "\n";
      continue;
    }
    if (n != 1 && isprime(n)) {
      c1++;
    } else
      c1 = 2;
    if (c == 1 && c1 == 1) {
      cout << "FastestFinger"
           << "\n";
    } else if (c == 1 && c1 > 1) {
      cout << "Ashishgup"
           << "\n";
    } else if (c == 1 && c1 == 0) {
      cout << "Ashishgup"
           << "\n";
    } else if (c1 == 0) {
      cout << "FastestFinger"
           << "\n";
    } else {
      cout << "Ashishgup"
           << "\n";
    }
  }
  return 0;
}

```