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:
from sys import stdin, stdout
from sys import maxsize
input = stdin.readline
# def print(n):
#    stdout.write(str(n)+'\n')


def solve():
    pass


test = 1
test = int(input().strip())
for t in range(0, test):
    n = int(input().strip())
    # s = list(input().strip())                                                        # String Input, converted to mutable list.
    # n, k = list(map(int, input().strip().split()))
    a = [int(x) for x in input().strip().split()]
    flag = True
    for i in range(n):
        if(a[i] != 1):
            flag = False
    if(flag):
        ans = n % 2
    else:
        i = 0
        while(a[i] == 1):
            i += 1
        ans = (i+1) % 2

    if(ans == 1):
        print('First')
    else:
        print('Second')
    # brr = [list(map(int,input().strip().split())) for i in range(rows)]              # 2D array row-wise input
    ans = solve()


'''
rows, cols = (5, 5)
arr = [[0]*cols for j in range(rows)]                                         # 2D array initialization
rev_str=s[::-1]                                                               # To reverse given string
b=input().strip().split()                                                     # list created by spliting about spaces
brr = [[int(b[cols*i+j]) for j in range(cols)] for i in range(rows)]          # 2D array Linear Input
rows,cols=len(brr),len(brr[0])                                                # no of rows/cols for 2D array
arr.sort(key = lambda x : x[1])                                               # sort list of tuples by 2nd element, Default priority - 1st Element then 2nd Element
s=set()                                                                       # empty set
a=maxsize                                                                     # initializing infinity
b=-maxsize                                                                    # initializing -infinity
mapped=list(map(function,input))                                              # to apply function to list element-wise
try:                                                                          # Error handling
    #code 1
except:                                                                       # ex. to stop at EOF
    #code 2 , if error occurs
'''