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. 
/* package codechef; // don't place package name! */

import java.lang.*;
import java.io.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		FastReader sc = new FastReader();
		PrintWriter out = new PrintWriter(System.out);
		int t = sc.nextInt();
		while(t-->0)
		{
		    int n = sc.nextInt();
		    boolean first=true;
		    int arr[] = new int[n];
		    for(int i=0;i<n;i++) arr[i]=sc.nextInt();
		    for(int i=0;i<n-1;i++)
		    {
		        if(arr[i]!=1) break;
		        else
		        {
		            if(first==true) first=false;
		            else first=true;
		        }
		    }
		    if(first) out.println("First");
		    else out.println("Second");
		}
		
		
		out.close();
	}
}

class FastReader 
{ 
    BufferedReader br; 
    StringTokenizer st; 
 
    public FastReader() 
    { 
        br = new BufferedReader(new
                 InputStreamReader(System.in)); 
    } 
 
    String next() 
    { 
        while (st == null || !st.hasMoreElements()) 
        { 
            try
            { 
                st = new StringTokenizer(br.readLine()); 
            } 
            catch (IOException  e) 
            { 
                e.printStackTrace(); 
            } 
        } 
        return st.nextToken(); 
    } 
 
    int nextInt() 
    { 
        return Integer.parseInt(next()); 
    } 
 
    long nextLong() 
    { 
        return Long.parseLong(next()); 
    } 
 
    double nextDouble() 
    { 
        return Double.parseDouble(next()); 
    } 
 
    String nextLine() 
    { 
        String str = ""; 
        try
        { 
            str = br.readLine(); 
        } 
        catch (IOException e) 
        { 
            e.printStackTrace(); 
        } 
        return str; 
    } 
}