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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
vector<long long> t;
void build(vector<long long> &a, long long v, long long tl, long long tr) {
  if (tl == tr) {
    t[v] = a[tl];
  } else {
    long long tm = (tl + tr) / 2;
    build(a, 2 * v, tl, tm);
    build(a, 2 * v + 1, tm + 1, tr);
    t[v] = t[2 * v] + t[2 * v + 1];
  }
}
void update(long long v, long long tl, long long tr, long long pos,
            long long value) {
  if (tl == tr) {
    t[v] = value;
  } else {
    long long tm = (tl + tr) / 2;
    if (pos <= tm) {
      update(2 * v, tl, tm, pos, value);
    } else {
      update(2 * v + 1, tm + 1, tr, pos, value);
    }
    t[v] = t[2 * v] + t[2 * v + 1];
  }
}
long long sum(long long v, long long tl, long long tr, long long l,
              long long r) {
  if (r < l) return 0;
  if (tl == l && tr == r) {
    return t[v];
  }
  long long tm = (tl + tr) / 2;
  return sum(2 * v, tl, tm, l, min(r, tm)) +
         sum(2 * v + 1, tm + 1, tr, max(l, tm + 1), r);
}
void SieveOfEratosthenes(long long n, bool *prime) {
  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;
    }
  }
}
long long min_(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
bool cycle = false;
vector<long long> topsort;
void dfs(long long u, vector<vector<long long>> &graph,
         vector<long long> &visited) {
  visited[u] = 1;
  for (long long i = 0; i < graph[u].size(); i++) {
    if (visited[graph[u][i]] == 1) {
      cycle = true;
    }
    if (visited[graph[u][i]] == 0) {
      dfs(graph[u][i], graph, visited);
    }
  }
  visited[u] = 2;
  topsort.push_back(u);
}
void primeFactors(long long n, map<long long, long long> &ma) {
  while (n % 2 == 0) {
    ma[2] = 1;
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      ma[i] = 1;
      n = n / i;
    }
  }
  if (n > 2) {
    ma[n] = 1;
  }
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long i, n, a, temp, ans = 0, ind, tc;
  cin >> tc;
  while (tc--) {
    cin >> n;
    if (n == 1) {
      cout << "FastestFinger\n";
      continue;
    }
    if (n == 2) {
      cout << "Ashishgup\n";
      continue;
    }
    if (n % 2 == 1) {
      cout << "Ashishgup\n";
      continue;
    }
    long long j = n;
    while (j % 2 == 0) {
      j /= 2;
    }
    if (isPrime(n / 2) || j == 1) {
      cout << "FastestFinger\n";
    } else {
      cout << "Ashishgup\n";
    }
  }
}
