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.Scanner;

public class SequentialNim {
    static Scanner sc;
    public static void main(String args[]){
        sc = new Scanner(System.in);
        int Case = sc.nextInt();
        for (int i=0;i<Case;i++){
            Solve();
        }
    }
    private static void Solve(){
        int n = sc.nextInt();
        int number1 = 0;
        boolean flag = false;
        int winner = 0;
        for (int i=0;i<n;i++){
            int tmp = sc.nextInt();
                if (flag){

                }else {
                    if (tmp==1){
                        number1++;
                    }else if (tmp>1){
                        flag = true;
                        winner=(number1+1)%2;
                    }
                }
        }
        if (!flag){
            winner = number1%2;
        }
        if (winner==0){
            System.out.println("Second");
        }else {
            System.out.println("First");
        }
    }
}
/***
 *
 *
 *
 *
 *   >1 == 1
 *   |1|%2 = 0 2
 *      %2 = 1 1
 *
 *     3 1 1 1 1 1
 *   1 3 1 1 1 1
 *   1> 1 0
 *
 *   1 2 2 1 1
 *     3 3 1 1 3 3 1 1
 *
 *
 *
 *
 *
 * */