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:
#
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#           ____          _     Generatered using
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#         | |    __ _  __| | ___ _ __  ______ _
#         | |   / _` |/ _` |/ _ \ '_ \|_  / _` |
#         | |__| (_| | (_| |  __/ | | |/ / (_| |
#          \____\____|\____|\___|_| |_/___\____|
#
#      GNU Affero General Public License v3.0
#    ------------------------------------------------
#    Author   : prophet
#    Created  : 2020-07-21 10:48:41.540377
#    UUID     : ueVtZeZVGsGHjX2O
#    ------------------------------------------------
#
production = True

import sys, math, collections

def input(input_format = 0, multi = 0):

    if multi > 0: return [input(input_format) for i in range(multi)]
    else:
        next_line = sys.stdin.readline()[:-1]

        if input_format >= 10:
            use_list = False
            input_format = int(str(input_format)[-1])
        else: use_list = True

        if input_format == 0: formatted_input = [next_line]
        elif input_format == 1: formatted_input = list(map(int, next_line.split()))
        elif input_format == 2: formatted_input = list(map(float, next_line.split()))
        elif input_format == 3: formatted_input = list(next_line)
        elif input_format == 4: formatted_input = list(map(int, list(next_line)))
        elif input_format == 5: formatted_input = next_line.split()
        else: formatted_input = [next_line]

        return formatted_input if use_list else formatted_input[0]

def out(output_line, output_format = 0, newline = True):

    formatted_output = ""

    if output_format == 0: formatted_output = str(output_line)
    elif output_format == 1: formatted_output = " ".join(map(str, output_line))
    elif output_format == 2: formatted_output = "\n".join(map(str, output_line))

    print(formatted_output, end = "\n" if newline else "")

def log(*args):
    if not production:
        print("$$$", end = "")
        print(*args)

enu = enumerate

ter = lambda a, b, c: b if a else c

ceil = lambda a, b: -(-a // b)

def mapl(iterable, format = 0):
    
    if format == 0: return list(map(int, iterable))
    elif format == 1: return list(map(str, iterable))
    elif format == 2: return list(map(list, iterable))
#
#   >>>>>>>>>>>>>>> START OF SOLUTION <<<<<<<<<<<<<<
#


def solve():

    n = input(11)
    a = input(1)

    t = 0

    for i, j in enu(a):
        if j != 1:
            out(ter(i & 1, "Second", "First"))
            return
    else:
        out(ter(n & 1, "First", "Second"))

    return


for i in range(input(11)): solve()
# solve()

#
#   >>>>>>>>>>>>>>>> END OF SOLUTION <<<<<<<<<<<<<<<
#