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.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;

public class Cookoff2 {

    public static ArrayList<Integer> list = new ArrayList<>();
    public static void main(String[] args) {
        int t;
        Scanner s = new Scanner(System.in);
        t=s.nextInt();
        while(t-->0)
        {
            int n=s.nextInt();
            ArrayList<Integer> arr = new ArrayList<>();
            int k;
            int sum=0;
            for (int i = 0; i <n ; i++) {
                k=s.nextInt();
                sum+=k;
                arr.add(k);
            }
            int temp=-1;
            for (int i = 0; i <n ; i++) {
                if(arr.get(i)>1)
                {
                    temp=i;
                    break;
                }
            }
            if(temp%2==0 && temp!=-1)
            {
                System.out.println("First");
            }
            else if (temp%2!=0 && temp!=-1)
            {
                System.out.println("Second");
            }
            else if(temp==-1)
            {
                if(n%2==0)
                {
                    System.out.println("Second");
                }
                else
                {
                    System.out.println("First");
                }
            }

        }
    }



}