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:
//Codeforces 1382B 
import java.util.Scanner;

public class CF1382B {
    static final Scanner SC = new Scanner(System.in);
    public static void main(String[] args) {
        StringBuilder solution = new StringBuilder();
        int tests = SC.nextInt();
        for (int t = 0; t < tests; ++t) {
            int stonePiles = SC.nextInt();
            int[] pile = new int[stonePiles];
            for (int p = 0; p < stonePiles; ++p)
                pile[p] = SC.nextInt();
            String winner = computeWinner(stonePiles, pile);
            solution.append(winner + "\n");
        }
        System.out.print(solution.toString());
    }

    // Computes and returns who the winner will be on the basis of the number of piles and the amount of stones in them
    static String computeWinner(int stonePiles, int[] pile) {
        int decidingIdx = -1;
        for (int p = 0; p < stonePiles; ++p)
            if (pile[p] != 1) {
                decidingIdx = p;
                break;
            }

        if (decidingIdx != -1) { // Some player finds a means to control the game hereafter
            return decidingIdx % 2 == 0 ? "First" : "Second";
        }
        else {
            return stonePiles % 2 != 0 ? "First" : "Second";
        }
    }
}