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.IOException;
import java.util.ArrayList;
import java.util.Scanner;
import java.util.Stack;

public class E {

	static int mod = (int) 1e9 + 7;

	static int ar[];
	static Scanner sc = new Scanner(System.in);
	static StringBuilder out = new StringBuilder();

	public static void main(String[] args) throws IOException {
		int t = sc.nextInt();

		while (t-- > 0) {
			int n=sc.nextInt();
			ar=new int[n];
			for(int i=0;i<n;i++) {
				ar[i]=sc.nextInt();
			}
			int win=n%2==0?2:1;
			for(int i=0;i<n;i++) {
				if(ar[i]!=1) {
					if(i%2==0) {
						win=1;
						break;
					}
					else {
						win=2;
						break;
						
					}
				}
				
			}
			out.append(win==2?"Second\n":"First\n");
		}
		System.out.println(out);
	}

	static int ab[] = new int[10000000];
	static int fans = 10000000;

	static void solve(int i, int n, int a[]) {
		if (i == n) {
			int cnt = 0;
			for (int j = 1; j < n; j++) {
				if (ar[j] < ar[j - 1]) {

					if (ab[j] >= ab[j - 1])
						cnt++;
				} else if (ar[j] > ar[j - 1]) {

					if (ab[j] <= ab[j - 1])
						cnt++;

				} else {
					if (ab[j] != ab[j - 1])
						cnt++;

				}

			}

			fans = Math.min(fans, cnt);
			return;
		}

		for (int k = 0; k < 4; k++) {
			ab[i] = k;
			solve(i + 1, n, a);
		}

	}

}