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 problemA 
{
   public static void main (String[] args) throws IOException
   {
   
      BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
      StringTokenizer token = new StringTokenizer(input.readLine());
      int maxCount = Integer.parseInt(token.nextToken());
      for(int b = 0; b < maxCount; b++)
      {
         boolean firstTurn = true;
         boolean secondTurn = false;
         boolean beginningOnes = true;
         token = new StringTokenizer(input.readLine());
         int maxLength = Integer.parseInt(token.nextToken());
         String[] array = input.readLine().split(" ");
         for(int x = 0; x < maxLength; x++)
         {
            if(x == maxLength - 1)
            {
               if(firstTurn)
                  System.out.println("First");
               else
                  System.out.println("Second");
            }
            else if(beginningOnes && Integer.parseInt(array[x]) == 1)
            {
               if(firstTurn)
               {
                  firstTurn = false;
                  secondTurn = true;
               }
               else
               {
                  firstTurn = true;
                  secondTurn = false;
               }
            }
            else
               beginningOnes = false;
         }
      }     
   }
}