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
import math,sys
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict
from os import path
import bisect as bi
import heapq 
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')): 
    #------------------Sublime--------------------------------------#
    sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
    def I():return (int(input()))
    def In():return(map(int,input().split()))
else:
    #------------------PYPY FAst I/o--------------------------------#
    def I():return (int(stdin.readline()))
    def In():return(map(int,stdin.readline().split()))

def dict(a):
    d={}
    for x in a:
        if d.get(x,-1)!=-1:
            d[x]+=1
        else:
            d[x]=1
    return d

def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bi.bisect_right(a, x)
    if i != len(a):
        return i
    else:
        return -1
def main():
    try:
        n=I()
        l=list(In())
        one=0
        pos=-1
        for x in range(n):
            if l[x]==1:
                one+=1
            if pos==-1:
                if l[x]!=1:
                    pos=x
        if one==n:
            if n&1==1:
                print('First')
            else:
                print('Second')
        else:
            if pos%2==0:
                print('First')
            else:
                print('Second')

    except:
        pass
        
M = 998244353
P = 1000000007
 
if __name__ == '__main__':
    for _ in range(I()):main()
    #for _ in range(1):main()