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:
import sys
import heapq, functools, collections
import math, random
from collections import Counter, defaultdict

# available on Google, not available on Codeforces
# import numpy as np
# import scipy


def solve(lst):  # fix inputs here
    console("----- solving ------")

    if set(lst) == set([1]):
        console("all ones")
        if len(lst)%2 == 0:
            return "Second"
        return "First"

    if lst[0] > 1:
        console("first choice")
        return "First"

    for i,x in enumerate(lst):
        if x > 1:
            if i%2 == 1:
                return "Second"
            return "First"

    return "Second"  # error

def console(*args):  # the judge will not read these print statement
    # print('\033[36m', *args, '\033[0m', file=sys.stderr)
    return

# fast read all
# sys.stdin.readlines()

for case_num in range(int(input())):
    # read line as a string
    # strr = input()

    # read line as an integer
    _ = int(input())
    
    # read one line and parse each word as a string
    # lst = input().split()

    # read one line and parse each word as an integer
    lst = list(map(int,input().split()))

    # read matrix and parse as integers (after reading read nrows)
    # lst = list(map(int,input().split()))
    # nrows = lst[0]  # index containing information, please change
    # grid = []
    # for _ in range(nrows):
    #     grid.append(list(map(int,input().split())))

    res = solve(lst)  # please change
    
    # Google - case number required
    # print("Case #{}: {}".format(case_num+1, res))

    # Codeforces - no case number required
    print(res)