This is the O(1) solution based on the pattern.

for _ in xrange(input()):
     print ["First","Second"][input()%7 in [0,1]]

The pattern emerges pretty quickly if you start from 1 stone and look at if the first player will win or lose with each stone you add.

SPOILER

You'll find that the first player will lose when the starting number of stones s is: 1,7,8,14,15,21,22... (i.e. when s % 7 < 2). With this knowledge in hand the code is trivial:

t = int(raw_input())
for _ in xrange(t):
    s = int(raw_input())
    if s % 7 < 2:
        print 'Second'
    else:
        print 'First'

Guys try it yourself for like 20 cases, on a paper! You will find out :P

This game is composed of the repeating group of LWWWWWL Therefore the solution in Java is like this --> System.out.println(scanner.nextInt()%7<2?"Second":"First");

Source: https://www.cs.umd.edu/~gasarch/COURSES/250/S15/nimnotes.pdf

C++ Solution and Explaination:

At first I had tried to modulos N by 10 (n = n%10) and set the number of if statements for each possibility from 1 to 10 to determine the winner, I was completely wrong. So I purchased the Test Cases for TC #0 and noticed that the pattern repeated itself after the 7th entry, it would go "Second First First First First First Second" and after that the pattern was the same. So I realized the best answer was to check if n was equal or greater than 7 and if so modulus n by 7 (n = n % 7). If the remainder was 0, it meant n was equal to the 7th case of the pattern (7 or 14 for example) and Player 2 Wins or if the reminder was 1, it meant n was equal to the 1st case of the pattern (8 or 15 for example) and Player 2 wins, any other remainder meant Player 1 wins.

int main() {
    int t;
    cin >> t;
    
    for(int i = 0; i < t; i++){
        int n;
        cin >> n;
        
        if(n >= 7)
            n %= 7;
        
        if(n == 0 || n == 1)
            cout << "Second" <<  endl;
        else
            cout << "First" << endl;
        
    }
    return 0;
}