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:
# Uses python3
from sys import stdin, stdout 
import sys
import math
import itertools
sys.setrecursionlimit(10**6)


def game(a,n):

	if a[0]!=1 or n==1:
		return "First"

	c=0
	for i in range(0,n):
		if a[i]==1:
			c+=1
		else:
			break

	if c==n:
		if c&1:
			return "First"
		else:
			return "Second"

	elif c&1:
		return "Second"
	else:
		return "First"



    
#python fast io

# suppose a function called main() and 
# all the operations are performed 
def main(): 
	t = int(stdin.readline().strip())
	for i in range (1,t+1):
		n = int(stdin.readline().strip())
		arr = [int(x) for x in stdin.readline().strip().split()]

		ans = game(arr,n) 
 

	# print answer via write 
	# write method writes only 
	# string operations 
	# so we need to convert any 
	# data into string for input 
		stdout.write(ans+'\n') 

# call the main method 
if __name__ == "__main__": 
	main()