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:
from math import ceil, log, floor, sqrt
import math	
	
	
k = 1	
def mod_expo(n, p, m):
	"""find (n^p)%m"""
	result = 1
	while p != 0:
		if p%2 == 1:
			result = (result * n)%m
		p //= 2
		n = (n * n)%m
	return result
	
def is_prime(n):
	m = 2
	while m*m <= n:
		if n%m == 0:
			return False
		m += 1
	return True
	
def find_sum(n, a):
	a.insert(0, 0)
	for i in range(1, n+1):
		prev = a[i] & a[i-1]
		cur = a[i] | a[i-1]
		a[i-1] = prev
		a[i] = cur
	return sum(m*m for m in a)

def prin_abc(x, y, z):
	l = [x, y, z]
	l.sort()
	if l[1] < l[2]:
		print("NO")
		return 
	a = b = l[0]
	c = l[2]
	print("YES")
	print(a, b, c)
	
def get_rank(n, stones):
	x = 0
	turn = 0
	winner = False
	for i in range(n-2, -1, -1):
		if winner:
			winner = not winner
		elif stones[i] == 1 and not winner:
			winner = True
		#print(winner)
		
	if (not winner) or (stones[0] > 1):
		print("First")
	else:
		print("Second")
t = 1
t = int(input())
while t:
	t = t - 1
	points = []
	n = int(input()) 
	#x, y, z = map(int, input().split()) 
	#print(discover())
	# = map(int, input().split())
	a = list(map(int, input().strip().split()))[:n]
	#w = list(map(int, input().strip().split()))[:k]
	#for i in range(3):
	#	x, y = map(int, input().split()) 
	#	points.append((x, y))
	#s = input()
	#if happy_guests(a, b, n, m):
#		print("Yes")
#	else:
#		print("No")
	
	#prin_abc(x, y, z)
	#print(find_sum(n, a))
	get_rank(n, a)