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. 
######################################################
############Created by Devesh Kumar###################
#############devesh1102@gmail.com####################
##########For CodeForces(Devesh1102)#################
#####################2020#############################
######################################################
import sys
input = sys.stdin.readline
 
# import sys
import heapq 
import copy
import math
import decimal
 
# import sys.stdout.flush as flush
# from decimal import *
#heapq.heapify(li) 
#
#heapq.heappush(li,4) 
#
#heapq.heappop(li)
#
# &    Bitwise AND Operator    10 & 7 = 2
# |    Bitwise OR Operator    10 | 7 = 15
# ^    Bitwise XOR Operator    10 ^ 7 = 13
 
# <<    Bitwise Left Shift operator    10<<2 = 40
# >>    Bitwise Right Shift Operator
# '''############ ---- Input Functions ---- #######Start#####'''
 
 
def inp():
    return(int(input()))
def inlt(): 
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def insr2():
    s = input()
    return((s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
############ ---- Input Functions ---- #######End
# #####   
 
 
 

def pr_list(a):
    print(*a, sep=" ")
def main():
    tests =  inp()
    # tests = 1
    mod = 1000000007
    limit = 10**18
    ans = 0
    stack = []
    hashm = {}
    arr = []
    heapq.heapify(arr) 

    for test in range(tests):
        n = inp()
        a = inlt()
        dp = [1 for i in range(n)]
        for i in range(n-2,-1,-1):
            if a[i] !=1:
                dp[i] = 1
            else:
                dp[i] = not(dp[i+1])
        if dp[0] == 1:
            print("First")

        else:
            print("Second")        
if __name__== "__main__":
    main()
    