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
from itertools import permutations, combinations
from collections import defaultdict
from collections import Counter
from bisect import bisect_left
import sys
from queue import PriorityQueue
import operator as op
from functools import reduce

mod = 1000000007


def L():
    return list(map(int, stdin.readline().split()))


def In():
    return map(int, stdin.readline().split())


def I():
    return int(stdin.readline())


def printIn(ob):
    return stdout.write(str(ob) + '\n')


def powerLL(n, p):
    result = 1
    while (p):
        if (p & 1):
            result = result * n % mod
        p = int(p / 2)
        n = n * n % mod
    return result


def ncr(n, r):
    r = min(r, n - r)
    numer = reduce(op.mul, range(n, n - r, -1), 1)
    denom = reduce(op.mul, range(1, r + 1), 1)
    return numer // denom


# --------------------------------------

def myCode():
    n = I()
    lst = L()
    if lst[0]!=1:
        print("First")
    elif lst.count(1)==len(lst):
        if n%2==0:
            print("Second")
        else:
            print("First")
    else:
        cnt = 0
        s = ''
        for i in lst:
            if i==1:
                cnt+=1
                if cnt%2==1:
                    s="Second"
                else:
                    s = "First"
            else:
                break
        print(s)
def main():
    for t in range(I()):
    	myCode()


if __name__ == '__main__':

    main()