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.*;
import java.util.*;

public class Solution
{
	
  	public static void main(String[] args) 
  	{	
		PrintWriter out=new PrintWriter(System.out);
		InputReader in=new InputReader(System.in);
		int t = in.nextInt();
		for (int ti = 0; ti < t; ti++) 
		{
			int n = in.nextInt();
			int firstNon = n+1;
			for (int i = 0; i < n; i++) 
			{
				int a = in.nextInt();
				if (a > 1) firstNon = Math.min(firstNon, i);
			}
			out.println(firstNon%2 == 0 ? "First" : "Second");
		}
				
    	out.close();
    }
	
	static class InputReader
	{
		BufferedReader br;
		StringTokenizer to;
		
		InputReader(InputStream stream)
		{
			br=new BufferedReader(new InputStreamReader(stream));
			to=new StringTokenizer("");
		}
		
		String next() 
		{
			while(!to.hasMoreTokens()) 
			{
				try
				{
				 	to=new StringTokenizer(br.readLine());
				}
				catch(IOException e) {}
			}
			return to.nextToken();
		}
		
		int nextInt() 
		{
			return Integer.parseInt(next());
		}
		
		long nextLong() 
		{
			return Long.parseLong(next());
		}
		
		int[] readArray(int n) 
		{
			int array[]=new int[n];
			for(int i=0; i<array.length; i++)
				array[i]=nextInt();
			return array;
		}
	}
	public static void printarray (int[] a, int n)
	{
		for(int i=0;i<n;i++)
		{
			System.out.print(a[i]+" ");
		}
	}	
	public static String strinsertcharacter (String s, int i, char c)
	{
		String str="";
		for(int j=0;j<i;j++)
		{
			str=str+s.charAt(j);
		}
		str=str+c;
		for(int j=i;j<s.length();j++)
		{
			str=str+s.charAt(j);
		}
		return str;
	}
	public static String strremoveele (String s, int i)
	{
		String str="";
		for(int j=0;j<s.length();j++)
		{
			if(j!=i)
			str=str+s.charAt(j);
		}
		return str;
	}
}