Solve the programming task below in a Python markdown code block.
Nim is the most famous two-player algorithm game. The basic rules for this game are as follows:

The game starts with a number of piles of stones.  The number of stones in each pile may not be equal.
The players alternately pick up $\mbox{1}$ or more stones from $\mbox{1}$ pile
The player to remove the last stone wins.

For example, there are $n=3$ piles of stones having $pile=[3,2,4]$ stones in them.  Play may proceed as follows:

Player  Takes           Leaving
                        pile=[3,2,4]
1       2 from pile[1]  pile=[3,4]
2       2 from pile[1]  pile=[3,2]
1       1 from pile[0]  pile=[2,2]
2       1 from pile[0]  pile=[1,2]
1       1 from pile[1]  pile=[1,1]
2       1 from pile[0]  pile=[0,1]
1       1 from pile[1]  WIN

Given the value of $n$ and the number of stones in each pile, determine the game's winner if both players play optimally.

Function Desctription  

Complete the nimGame function in the editor below.  It should return a string, either First or Second.  

nimGame has the following parameter(s):  

pile: an integer array that represents the number of stones in each pile  

Input Format

The first line contains an integer, $\mathrm{~g~}$, denoting the number of games they play.

Each of the next $\mathrm{~g~}$ pairs of lines is as follows:  

The first line contains an integer $n$, the number of piles.
The next line contains $n$ space-separated integers $pile[i]$, the number of stones in each pile.

Constraints

$1\leq g\leq100$
$1\leq n\leq100$
$0\leq s_i\leq100$
Player 1 always goes first.

Output Format

For each game, print the name of the winner on a new line (i.e., either First or Second).

Sample Input
2
2
1 1
3
2 1 4

Sample Output
Second
First

Explanation

In the first case, there are $n=2$ piles of $pile=[1,1]$ stones.  Player $\mbox{1}$ has to remove one pile on the first move.  Player $2$ removes the second for a win.

In the second case, there are $n=3$ piles of $pile=[2,1,4]$ stones.  If player $\mbox{1}$ removes any one pile, player $2$ can remove all but one of another pile and force a win.  If player $\mbox{1}$ removes less than a pile, in any case, player $2$ can force a win as well, given optimal play.