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:


# template by 3xC and starkizard.
# contributors: 
    
#####################################################################################
from __future__ import division, print_function 
    
import sys
import os
from collections import Counter, deque, defaultdict
import itertools
import math
import io
    
    
"""Uncomment modules according to your need"""
    
    
# from bisect import bisect_left, bisect_right, insort
# from heapq import heappop, heapify, heappush
# from random import randint as rn
# from Queue import Queue as Q
# from copy import deepcopy
# from decimal import *
# import re
# import operator
    
#####################################################################################
# this enables you to write python3 code with PyPy2 (Python 2)
if sys.version_info[0] < 3:
    input = raw_input
    range = xrange
    
    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip
    
#####################################################################################
    
"""value of mod"""
MOD = 998244353
mod = 10**9 + 7
    
"""Uncomment next 4 lines if doing huge recursion"""
# import threading
# threading.stack_size(1<<27)
# sys.setrecursionlimit(10000
    
def prepare_factorial(mod=mod):
    """ returns two lists, factorial and inverse factorial modulo argument by default 10**9 +7 """
    # Comment code out when you don't need inverse factorial or vice versa
    fact = [1]
    for i in range(1, 200005):
        fact.append((fact[-1] * i) % mod)
    
    ifact = [0] * 200005
    ifact[200004] = pow(fact[200004], mod - 2, mod)
    for i in range(200004, 0, -1):
        ifact[i - 1] = (i * ifact[i]) % mod
    
    return fact, ifact
    
def modinv(n, p):
    """ returns N inverse modulo p """
    return pow(n, p - 2, p)
    
def ncr(n, r,  fact, ifact):
    """ takes 4 arguments: n , r and factorial and inverse factorial lists"""
    t = (fact[n] * (ifact[r]*ifact[n-r]) % MOD)% MOD
    return t
    
    
def get_n(Sum):    
    """this function returns the maximum n for which Summation(n) <= Sum"""
    ans = (-1 + sqrt(1 + 8*Sum))//2
    return ans
    
def sieve(n):
    """ returns a list of prime numbers till n """
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r
    
    
    
def divs(n, start=1):
    """ returns a list of all divisors till n """
    divisors = []
    #rdivisors=[]
    for i in range(start, int(math.sqrt(n) + 1)):
        if n % i == 0:
            if n / i == i:
                divisors.append(i)
            else:
                divisors.extend([i, n // i])
    return divisors
    
    
    
def divn(N):
    """ returns the number of divisors """
    divs_number = 1
    factorCounter=1
    tempList=[]
    
    while N%2==0:
        factorCounter+=1
        N=N//2
    tempList.append(factorCounter)
    
    i=3
    
    while i*i<=N:
        factorCounter=1
        while N%i==0:
            factorCounter+=1
            N=N//i
        if factorCounter!=1:
            tempList.append(factorCounter)
        i+=1
    
    if N>2:
        factorCounter=2
        tempList.append(factorCounter)
    
    divisorCount=1
    for i in tempList:
        divisorCount*=i
    return(divisorCount)
    
    
    
def lrfind(d, x, default=-1):
    """ Takes 2 arguments an iterable and an element. returns a tuple (firstoccurence,lastoccurence) -1 if not found """
    left = right = -1
    for i in range(len(d)):
        if d[i] == x:
            if left == -1: left = i
            right = i
    if left == -1:
        return default, default
    else:
        return left, right
    
def gcd(x, y): # math.gcd is slower
    """ returns greatest common divisor of x and y """
    while y:
        x, y = y, x % y
    return x
    
def check_sorted(a):
    ''' returns True/False '''
    for i in range(len(a)-1):
        if a[i]>a[i+1]:
            return False
    return True
def ceil(n, k=1): return n // k + (n % k != 0) #returns math.ceil but protecting against floating inconsistencies
def input(): return sys.stdin.readline().strip()
def ii(): return int(input()) #inputs integer
def mi(): return map(int, input().split()) # inputting space seperated variables for example x,y,z
def li(): return list(map(int, input().split())) #inputting a space seperated list of integers
def lw(): return input().split() #inputting a space seperated list of strings
def lcm(a, b): return abs(a * b) // gcd(a, b) #returns LCM of two arguments
def prr(a, sep=' ', end='\n'): print(sep.join(map(str, a)), end=end) #For printing an iterable with seperator sep as optional second argument (default : " "), ending character (default: "\n") as optional third
def dd(): return defaultdict(int) #returns a dictionary with values defaulted to 0
def ddl(): return defaultdict(list) #returns a dictionary with values defaulted to []
def write(s): return sys.stdout.write(s)
    
    
    
    
    
    
    
    
    
    
    
    
###################################################################
    
def main():
    #CODE GOES HERE:
    for t in range(ii()):
        n=ii()
        a=li()
        if n==1:
            print("First")
            continue
        forced=0
        for i in a:
            if i==1:
                forced+=1
            else:
                break
        if forced==n:
            print(["Second","First"][n&1])
            continue
        print(["First","Second"][forced&1])
        

        
            
    
    
""" -------- Python 2 and 3 footer by Pajenegod and c1729 ---------"""
    
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
    
import os, sys
from io import IOBase, BytesIO
    
BUFSIZE = 8192
    
class FastIO(BytesIO):
    newlines = 0
    
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
    
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s
    
    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()
    
    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
    
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
    
    
class IOWrapper(IOBase):
    
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')
    
    
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
    
""" main function"""
    
if __name__ == '__main__':
    main()
    # threading.Thread(target=main).start()