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:
z=input
mod = 10**9 + 7
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
def lcd(xnum1,xnum2):
    return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
        p=ceil(x**.5)+1
        for i in range(2,p):
            if x%i==0:
                return 0
        return 1

################################################################################

"""

n=int(z())

for _ in range(int(z())):

x=int(z())

l=list(map(int,z().split()))

n=int(z())

l=sorted(list(map(int,z().split())))[::-1]

a,b=map(int,z().split())

l=set(map(int,z().split()))

led=(6,2,5,5,4,5,6,3,7,6)

vowel={'a':0,'e':0,'i':0,'o':0,'u':0}

color4=["G", "GB", "YGB", "YGBI", "OYGBI" ,"OYGBIV",'ROYGBIV' ]

"""

###########################---START-CODING---###############################################
num=int(z())

for _ in range(num):
    n=int(z())
    l1=list(map(int,z().split()))
    one=l1.count(1)

    if one==n:
        if one%2==0:
            print('Second')
        else:
            print('First')
        continue
    c=0
    for i in l1:
        if i==1:
            c+=1
        else:
            break
    if c%2==0:
        print('First')
    else:
        print('Second')