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>
#pragma GCC optimize("-Ofast", "-funroll-all-loops")
using namespace std;
const int N = 100010;
map<int, int> dp;
int dfs(int x) {
  if (dp.count(x)) return dp[x];
  if (x & 1) return 1;
  int ok = 0;
  for (int i = 2; i * i <= x && !ok; i++)
    if (x % i == 0) {
      if (i & 1) {
        if (!dfs(x / i)) ok = 1;
      }
      if ((x / i) & 1) {
        if (!dfs(i)) ok = 1;
      }
    }
  dp[x] = ok;
  return dp[x];
}
inline void solve() {
  int n;
  cin >> n;
  puts(dfs(n) ? "Ashishgup" : "FastestFinger");
}
signed main() {
  dp[1] = 0;
  dp[2] = 1;
  dp[3] = 1;
  for (int i = 5; i <= 100000; i++) {
    if (i & 1)
      dp[i] = 1;
    else {
      int ok = 0;
      for (int j = 2; j * j <= i; j++)
        if (i % j == 0) {
          if (j & 1) {
            if (!dp[i / j]) ok = 1;
          }
          if ((i / j) & 1) {
            if (!dp[j]) ok = 1;
          }
        }
      dp[i] = ok;
    }
  }
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}
