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. 
import sys
input = sys.stdin.buffer.readline
def I(): return(list(map(int,input().split())))
def sieve(n):
	a=[1]*n
	for i in range(2,n):
	    if a[i]:
	        for j in range(i*i,n,i):
	            a[j]=0
	return a
def rec(i,p):
	if dp[i][p]!=-1:
		return dp[i][p]
	# print(arr[i])
	if i==n-1:
		dp[i][p]=p
		return p
	if arr[i]==1:
		dp[i][p] =rec(i+1,1-p)
		# print("Yes",i,p,dp[i][p])
		return dp[i][p]
	p1 = rec(i+1,1-p)
	p2 = rec(i+1,p)
	if p1==p or p2==p:
		dp[i][p] = p
	else:
		dp[i][p] = 1-p
	return dp[i][p]


for __ in range(int(input())):
	n=int(input())
	arr =I()
	dp =[[-1,-1] for i in range(n)]
	# print(arr)
	for i in reversed(range(n)):
		rec(i,0)
		rec(i,1)
		# print(dp[i][0],dp[i][1],i,"###",arr[i])
	# print(dp)
	ans = rec(0,1)
	# g = n-arr.count(1)
	# ans  = True if g%2 else False
	# # ans = True if n%2 else False
	# # for i in range(n):
	# # 	if arr[i]>1:
	# # 		ans = not ans
	if ans==1:
		print("First")
	else:
		print("Second")



