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:
# def return_winner(a,int_winner):
# 	if a[0]!=1:
# 		print('First')
# 	else:
# 		b = a[:-1]
# 		b = b[1:]
# 		print(b)
# 		if 1 in b:
# 			c = b.index(1)
# 			print(c)
# 			if list(set(b[:c+1]))!=[1]:
# 				print('Second')
# 			else:
# 				print('First')
# 		else:
# 			print('Second')	
# 	pass


t = int(input())
for t1 in range(t):
	n = int(input())
	a = [int(n1) for n1 in input().split()]
	count = 0
	for i in range(len(a)):
		if a[i]!=1:
			count = 1 
			if i%2==0:
				print('First')
				break
			else:
				print('Second')
				break
		else:
			pass
	if count==0:
		if n%2==0:
			print('Second')
		else:
			print('First')
	# if list(set(a))==[1]:
	# 	if len(a)%2==0:
	# 		print('Second')
	# 	else:
	# 		print('First')
	# elif a[0]!=1:
	# 	print('First')
	# else:
	# 	b = a[:-1]
	# 	b = b[1:]
	# 	print(b)
	# 	if 1 in b:
	# 		c = b.index(1)
	# 		print(c)
	# 		if list(set(b[:c+1]))!=[1]:
	# 			print('Second')
	# 		else:
	# 			print('First')
	# 	else:
	# 		print('Second')