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.*;
import java.lang.Math;

//System.out.println();


public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int mod = 1000000007;

        for (int i = 0; i < n; i++) {
        	int k = sc.nextInt();
        	int sum = 0;
        	boolean flag= false;
        	boolean winner=false;
        	for (int j=0;j<k;j++) {
        		int temp = sc.nextInt();
        		if (temp == 1) {
        			if (!flag) {
        				winner = !winner;
        				continue;
        			} else {
        				sum++;
        			}
        		} else {
        			flag = true;

        		}
        	}

        	if (!flag) {
        		if (winner) {
        			System.out.println("First");
        		} else {
        			System.out.println("Second");
        		}
        	} else {
        		if (sum == 0) {
        			if (winner) {
        				System.out.println("Second");
        			} else {
        				System.out.println("First");

        			}
        		} else {
					if (sum % 2 == 1) {
						if (winner) {
							System.out.println("Second");
						} else {
							System.out.println("First");

						}
	        		} else {
	        			if (winner) {
							System.out.println("Second");
						} else {
							System.out.println("First");

						}

	        		}
        		}
        		
        	}
        }
        
    }
}