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;
const double EPS = 1e-9;
const double PI = 3.14159265359;
vector<long long> primes;
void GeneratingPrimes(int top) {
  vector<bool> p(top + 1, 1);
  p[0] = p[1] = 0;
  for (int i = 2; i <= sqrt(top); i++) {
    if (p[i]) {
      for (int j = i * i; j <= top; j += i) {
        p[j] = 0;
      }
    }
  }
  for (int i = 0; i < top + 1; i++)
    if (p[i]) primes.push_back(i);
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void EXTgcd(long long a, long long b, long long& x0, long long& y0) {
  if (b == 0) {
    x0 = 1, y0 = 0;
    return;
  }
  EXTgcd(b, a % b, x0, y0);
  long long x1 = y0;
  long long y1 = x0 - (a / b) * (y0);
  y1 = y1;
  x0 = x1;
  y0 = y1;
}
long long power(long long n, long long p, long long mod) {
  if (!p) return 1;
  long long ans = 1;
  while (p > 1) {
    if (p & 1) ans = (ans * n) % mod;
    n = (n * n) % mod;
    p /= 2;
  }
  return (ans * n) % mod;
}
double dist(double a, double b, double x, double y) {
  return sqrt(pow(a - x, 2) + pow(b - y, 2));
}
double dist3(double a, double b, double c, double x, double y, double z) {
  return sqrt(pow(a - x, 2) + pow(b - y, 2) + pow(c - z, 2));
}
int xadd[9] = {1, -1, 0, 1, -1, 0, 1, -1, 0};
int yadd[9] = {1, -1, 0, -1, 0, 1, 0, 1, -1};
int xx[4] = {0, 1, 0, -1};
int yy[4] = {1, 0, -1, 0};
long long const N = 1e5 + 10;
long long const INF = 1e18;
long long const mod = 998244353;
int t, n;
string p[2] = {"Ashishgup", "FastestFinger"};
int main() {
  GeneratingPrimes(1e6);
  cin >> t;
  while (t--) {
    cin >> n;
    if (n == 1) {
      cout << p[1] << endl;
      continue;
    }
    if (n & 1 || n == 2) {
      cout << p[0] << endl;
      continue;
    }
    int o = 0, two = 0;
    for (int pr : primes) {
      if (pr > n) break;
      while (n % pr == 0) {
        n /= pr;
        if (pr == 2)
          two++;
        else
          o++;
      }
    }
    if (n > 1) o++;
    if ((two > 1 && o) || o >= 2) {
      cout << p[0] << endl;
      continue;
    } else {
      cout << p[1] << endl;
      continue;
    }
  }
  return 0;
}
