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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
 
public class sequentialnim {
 
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		 int c = Integer.parseInt(in.readLine());
		
		for(int cases=0; cases<c; cases++) {
		
			
			int len = Integer.parseInt(in.readLine());
			StringTokenizer st = new StringTokenizer(in.readLine(), " ");
			int[] nums = new int[len];
			for(int i=0; i<len; i++) { //read input
				nums[i]=Integer.parseInt(st.nextToken());
			}
			int playin = 0;
			boolean isone = true;
			int countones = 0;
			while(isone && playin<len) { //run through beginning of array until we reach a non-1 number
				if(nums[playin]!=1) {
					isone=false; //if we run into a non-1 number
					break;
				}
				playin++; //index
				countones++; //number of ones
			}
			
			if(countones==len && len%2==0) { //if the number of ones in the array is even, and the whole array is 1's
				System.out.println("Second");
			}
			else if(countones==len && len%2!=0) { //if the number of ones in the array is odd, and the whole array is 1's
				System.out.println("First");
			}
			else if(countones%2==0) { //if the number of ones in the array is even
				System.out.println("First");
				
			}
			else if(countones%2!=0) { //if the number of ones in the array is odd
				System.out.println("Second");
			}
			
			else { //remember that first always wins except for the conditions above, considering they play optimally.
				System.out.println("First");
			}
		}
		in.close();
		out.close();
	}
 
}