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 sys import stdin, stdout
cin = stdin.readline
cout = stdout.write
mp = lambda:list(map(int, cin().split()))

t, = mp()
'''
for _ in range(t):
	n, m = mp()
	a = mp()
	b = mp()
	
	l1 = [0]* 10001
	l2 = [0]* 10001
	
	for i in range(n):
		l1[a[i]] = 1
	for i in range(m):
		l2[b[i]] = 1
		if l2[b[i]] + l1[b[i]] > 1:
			cout('YES\n')
			cout('1 ' + str(b[i]) + '\n')
			break
	else:
		cout('NO\n')
'''
for _ in range(t):
	n, = mp()
	a = mp()
	
	if sum(a) == n:
		if n%2:
			cout('First\n')
		else:
			cout('Second\n')
		continue
	f1 = 1
	for i in range(n-1):
		if a[i] == 1:
			f1 = (f1==1)*2 + (f1==2)*1
		else:
			if f1 == 2:
				cout('Second\n')
				break
			else:
				cout('First\n')
				break
	else:
		if f1 == 2:
			cout('Second\n')
				
		else:
			cout('First\n')