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 PYTHON3):

Solution:
import threading, sys

threading.stack_size(2**26)


sys.setrecursionlimit(10**6)


def main():
	a = []

	dp = []

	def recur(i, c):
		# print(i, c)

		if i == len(a)-1:
			dp[c][i] = c
			return c

		if dp[c][i] != -1:
			return dp[c][i]

		if a[i] == 1:
			ans = recur(i+1, 1-c)

		else:
			ans1 = recur(i+1, c)
			ans2 = recur(i+1, 1-c)

			if ans1 == ans2:
				ans = ans1

			else:
				ans  = c

		dp[c][i] = ans
		return ans




	t = int(input())

	for ii in range(t):

		n = int(input())

		a = list(map(int, input().split()))

		if len(a) == 1:
			print("First")
			continue

		if len(a) == 2:
			if a[0] == 1:
				print("Second")
			else:
				print("First")

			continue

		greater = False
		one = False

		for i in a:
			if i > 1:
				greater = True

			if i == 1:
				one = True

		if greater and not one:
			print("First")
			continue

		if not greater and one:
			if n%2 == 0:
				print("Second")

			else:
				print("First")
			continue

		dp = [[], []]

		for i in range(n):
			dp[0].append(-1)
			dp[1].append(-1)

		ans = recur(0, 1)

		if ans == 1:
			print("First")

		else:
			print("Second")

threading.Thread(target=main).start()