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:
'''      ___           ___                         ___                       ___           ___                         ___     
     /\__\         /\  \         _____         /\  \                     /\  \         /\  \                       /\__\    
    /:/ _/_        \:\  \       /::\  \        \:\  \       ___         /::\  \       |::\  \         ___         /:/ _/_   
   /:/ /\  \        \:\  \     /:/\:\  \        \:\  \     /\__\       /:/\:\__\      |:|:\  \       /\__\       /:/ /\  \  
  /:/ /::\  \   ___  \:\  \   /:/  \:\__\   ___ /::\  \   /:/__/      /:/ /:/  /    __|:|\:\  \     /:/  /      /:/ /::\  \ 
 /:/_/:/\:\__\ /\  \  \:\__\ /:/__/ \:|__| /\  /:/\:\__\ /::\  \     /:/_/:/__/___ /::::|_\:\__\   /:/__/      /:/_/:/\:\__\
 \:\/:/ /:/  / \:\  \ /:/  / \:\  \ /:/  / \:\/:/  \/__/ \/\:\  \__  \:\/:::::/  / \:\~~\  \/__/  /::\  \      \:\/:/ /:/  /
  \::/ /:/  /   \:\  /:/  /   \:\  /:/  /   \::/__/       ~~\:\/\__\  \::/~~/~~~~   \:\  \       /:/\:\  \      \::/ /:/  / 
   \/_/:/  /     \:\/:/  /     \:\/:/  /     \:\  \          \::/  /   \:\~~\        \:\  \      \/__\:\  \      \/_/:/  /  
     /:/  /       \::/  /       \::/  /       \:\__\         /:/  /     \:\__\        \:\__\          \:\__\       /:/  /   
     \/__/         \/__/         \/__/         \/__/         \/__/       \/__/         \/__/           \/__/       \/__/    
'''
"""
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"""
import sys
import math
import collections
import operator as op
from collections import deque
from math import gcd, inf, sqrt, pi, cos, sin, ceil, log2
from bisect import bisect_right, bisect_left

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

from functools import reduce
from sys import stdin, stdout, setrecursionlimit
setrecursionlimit(2**20)


def factorial(n):
    if n == 0:
        return 1
    return n * factorial(n - 1)


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  # or / in Python 2


def prime_factors(n):
    i = 2
    factors = []
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)
    if n > 1:
        factors.append(n)
    return (list(factors))


def isPowerOfTwo(x):
    return (x and (not(x & (x - 1))))


def factors(n):
    return list(set(reduce(list.__add__, ([i, n // i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))

MOD = 1000000007
PMOD = 998244353
T = 1
T = int(stdin.readline())
for _ in range(T):
    # n, m = list(map(int, stdin.readline().rstrip().split()))
    n = int(stdin.readline())
    a = list(map(int, stdin.readline().rstrip().split()))
    # b = list(map(int, stdin.readline().rstrip().split()))
    # q = int(stdin.readline())
    # c = list(map(int, stdin.readline().rstrip().split()))
    # a = str(stdin.readline().strip('\n'))
    ans = 1
    if n == sum(a):
        if n % 2 == 0:
            print('Second')
        else:
            print('First')
        continue

    for i in range(n):
        if a[i] != 1:
            break
        else:
            ans ^= 1
    if ans == 1:
        print('First')
    else:
        print('Second')