Problem: There are n piles of stones, where the i-th pile has a_i stones. Two people play a game, where they take alternating turns removing stones.

In a move, a player may remove a positive number of stones from the first non-empty pile (the pile with the minimal index, that has at least one stone). The first player who cannot make a move (because all piles are empty) loses the game. If both players play optimally, determine the winner of the game.

Input

The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 2t lines contain descriptions of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the number of piles.

The second line of each test case contains n integers a_1,…,a_n (1≤ a_i≤ 10^9) — a_i is equal to the number of stones in the i-th pile.

It is guaranteed that the sum of n for all test cases does not exceed 10^5.

Output

For each test case, if the player who makes the first move will win, output "First". Otherwise, output "Second".

Example

Input


7
3
2 5 4
8
1 1 1 1 1 1 1 1
6
1 2 3 4 5 6
6
1 1 2 1 2 2
1
1000000000
5
1 2 2 1 1
3
1 1 1


Output


First
Second
Second
First
First
Second
First

Note

In the first test case, the first player will win the game. His winning strategy is: 

  1. The first player should take the stones from the first pile. He will take 1 stone. The numbers of stones in piles will be [1, 5, 4]. 
  2. The second player should take the stones from the first pile. He will take 1 stone because he can't take any other number of stones. The numbers of stones in piles will be [0, 5, 4]. 
  3. The first player should take the stones from the second pile because the first pile is empty. He will take 4 stones. The numbers of stones in piles will be [0, 1, 4]. 
  4. The second player should take the stones from the second pile because the first pile is empty. He will take 1 stone because he can't take any other number of stones. The numbers of stones in piles will be [0, 0, 4]. 
  5. The first player should take the stones from the third pile because the first and second piles are empty. He will take 4 stones. The numbers of stones in piles will be [0, 0, 0]. 
  6. The second player will lose the game because all piles will be empty. 

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class B_Round658 {
	static int n;
	public static void main(String[] args) throws NumberFormatException, IOException
	{
		BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
		int tt = Integer.parseInt(scan.readLine());
		
		for(int t = 0; t < tt; t++)
		{
			n = Integer.parseInt(scan.readLine());
			String[] in = scan.readLine().split(" ");
			int[] ray = new int[n];
			int[][] winner = new int[n][2];
			
			
			for(int i = 0; i < n; i++)
			{
				ray[i] = parse(in[i]);
				winner[i][0] = -1;
				winner[i][1] = -1;
			}
			
			int win = goThrough(ray, winner, 0, 0);
			
			System.out.println(win == 0 ? "First" : "Second");
			
		}
	}

	private static int goThrough(int[] ray, int[][] winner, int index, int player) 
	{
		if(winner[index][player] != -1)
			return winner[index][player];
		
		if(index == n-1)
			return player;
		
		if(ray[index] == 1)
		{
			winner[index][player] = goThrough(ray, winner, index+1, (player+1)%2);
			return winner[index][player];
		}
		
		else
		{
			int temp = goThrough(ray, winner, index+1, player);
			if(temp != player)
				temp = goThrough(ray, winner, index+1,(player+1) % 2 );
			
			winner[index][player] = temp;
			return temp;
		}
	}

	public static int parse(String num)
	{
		return Integer.parseInt(num);
	}
}