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 PYTHON):

Solution:
from __future__ import division, print_function
_interactive = False

def main():
    for _ in range(int(input())):
        n = int(input())
        ar = input_as_list()

        ar[-1] = 2
        
        state = 0
        for e in ar:
            if e == 1:
                state = 1-state
            else:
                break

        print(['First', 'Second'][state])


# Constants
INF = float('inf')
MOD = 10**9+7
alphabets = 'abcdefghijklmnopqrstuvwxyz'

# Python3 equivalent names
import os, sys, itertools
if sys.version_info[0] < 3:
    input = raw_input
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip

# print-flush in interactive problems
if _interactive:
    flush = sys.stdout.flush
    def printf(*args, **kwargs):
        print(*args, **kwargs)
        flush()

# Debug print, only works on local machine
LOCAL = "LOCAL_" in os.environ
debug_print = (print) if LOCAL else (lambda *x, **y: None)

# Fast IO
if (not LOCAL) and (not _interactive):
    from io import BytesIO
    from atexit import register
    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
    sys.stdout = BytesIO()
    register(lambda: os.write(1, sys.stdout.getvalue()))
    input = lambda: sys.stdin.readline().rstrip('\r\n')

# Some utility functions(Input, N-dimensional lists, ...)
def input_as_list():
    return [int(x) for x in input().split()]

def input_with_offset(o):
    return [int(x)+o for x in input().split()]

def input_as_matrix(n, m):
    return [input_as_list() for _ in range(n)]

def array_of(f, *dim):
    return [array_of(f, *dim[1:]) for _ in range(dim[0])] if dim else f()

main()