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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;
import java.util.Stack;

public class Test {

	public static void main(String[] args) {

		Scanner scn = new Scanner(System.in);

		int t = scn.nextInt();

		while (t-- > 0) {
			int n = scn.nextInt();

			int a = -1, b = 10;

			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = scn.nextInt();
			}

			solve(arr, n);
		}

	}

	private static void solve(int[] array, int size) {
		int test = 0;
		int c = 0;
		int checkCondition = 0;
		long ans = 0;
		boolean flag = false;

		int ab = 0;
		int ID = -1;
		for (int i = 0; i < array.length; i++) {
			if (array[i] == 1) {
				c++;
			}

			if (array[i] != 1 && ID == -1) {
				ID = i;
			}

			if (checkCondition == 0 && array[i] > 1 && ab == 0) {
				test = 1;
			}

			if (array[i] > 1) {
				ab = 1;
			}

			checkCondition = checkCondition ^ 1;
		}

		if (c == size) {
			if (size % 2 == 1) {
				System.out.println("First");
				return;
			} else {
				System.out.println("Second");
				return;
			}
		}

		if (test == 1) {
			System.out.println("First");
		} else {
			System.out.println("Second");
		}
	}

}