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:
# @author --> ajaymodi
# Naive approach
import sys
# sys.stdin=open("input.in","r")
# sys.stdout=open("output.out","w")

input=lambda : sys.stdin.readline().strip()
char = [chr(i) for i in range(97,123)]
CHAR = [chr(i) for i in range(65,91)]
mp = lambda:list(map(int,input().split()))
INT = lambda:int(input())
rn = lambda:range(INT())

from math import ceil,sqrt,factorial,gcd

for _ in rn():
	n = INT()
	l = mp()

	count_1=0

	for i in range(n):
		if l[i]>1:
			if count_1%2==1:
				second=1
				first=0
			else:
				second=0
				first=1
			count_1=""
			break
		else:
			count_1+=1

	if count_1!="":
		if count_1%2==0:
			second=1
			first=0
		else:
			first=1
			second=0

	if first==1:
		print("First")
	else:
		print("Second")