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.util.ArrayList;
import java.util.StringTokenizer;
import java.io.InputStreamReader;
import java.io.IOException;
 
public class codeforces {
	
	public static void main(String[] args) {	
         scanner in  = new scanner();
         int t = in.nextInt();
         for(int i = 0;i<t;i++)
         {
        	 int n = in.nextInt();
        	 boolean one = true; 
        	 int count = 0;
        	 int other = 0;
        	for(int j = 0;j<n;j++)
        	{
        		int x = in.nextInt();
        		if(x!=1)
        		{
        			one = false;
        			other++;
        		}
        		if(one)
        			count++;	
        	}
        	if(other>0)
        	{
        		if(count%2!=0)
        			System.out.println("Second");
        		else
        			System.out.println("First");
        	}
        	else
        	{
        		if(count%2!=0)
        			System.out.println("First");
        		else
        			System.out.println("Second");
        	}
         }
	}
}
class scanner
{
	BufferedReader br ;
	StringTokenizer st;
	
	public scanner()
	{
		br = new BufferedReader(new InputStreamReader(System.in));
		st = null;
	}
	 public String next() {
	        while (st == null || !st.hasMoreElements()) {
	            try {
	                st = new StringTokenizer(br.readLine());
	            } catch (IOException e) {
	                e.printStackTrace();
	            }
	        }
	        return st.nextToken();
	    }
	public int nextInt()
	{
		return Integer.parseInt(next());
	}
	public long nextLong()
	{
	  return	Long.parseLong(next());
	}
}