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 math
from decimal import Decimal
import heapq
import copy
import heapq
from collections import deque
def na():
	n = int(input())
	b = [int(x) for x in input().split()]
	return n,b
 
	
def nab():
	n = int(input())
	b = [int(x) for x in input().split()]
	c = [int(x) for x in input().split()]
	return n,b,c
 
		
def dv():
	n, m = map(int, input().split())
	return n,m
 
 
def da():
	n, m = map(int, input().split())
	a = list(map(int, input().split()))
	return n,m, a 
 
 
def dva():
	
	n, m = map(int, input().split())
	a = [int(x) for x in input().split()]
	b = [int(x) for x in input().split()]
	return n,m,b
 
 
def eratosthenes(n): 
	sieve = list(range(n + 1))
	for i in sieve:
		if i > 1:
			for j in range(i + i, len(sieve), i):
				sieve[j] = 0
	return sorted(set(sieve))
 
 
def lol(lst,k):
	k=k%len(lst)
	ret=[0]*len(lst)
	for i in range(len(lst)):
		if i+k<len(lst) and i+k>=0:
			ret[i]=lst[i+k]
		if i+k>=len(lst):
			ret[i]=lst[i+k-len(lst)]
		if i+k<0:
			ret[i]=lst[i+k+len(lst)]
	return(ret)
def nm():
	n = int(input())
	b = [int(x) for x in input().split()]
	m = int(input())
	c = [int(x) for x in input().split()]
	return n,b,m,c
 
 
def dvs():
	n = int(input())
	m = int(input())
	return n, m 
 
 
def fact(n):
	tc = []
	ans = {}
	d = 2
	while d * d <= n:
		if n % d == 0:
			tc.append(d)
			n //= d
		else:
			d += 1
	if n > 1:
		tc.append(n)
	for i in tc:
		ans[i] = ans.get(i, 0) + 1
	return ans


def how_many(l, r, c):
	m = (l + r) // 2
	if r - l == 1:
		if s[l] == c:
			return 0
		else:
			return 1
	d = s[l:m]
	k1 = (r - l) // 2 - s[l:m].count(c) + how_many(m, r, chr(ord(c) + 1))
	k2 = (r - l) // 2 - s[m:r].count(c) + how_many(l, m, chr(ord(c) + 1))
	return min(k1, k2)
 
for _ in range(int(input())):
		n = int(input())
		a = list(map(int, input().split()))
		k = a.count(1)
		if k == n:
			if n % 2 == 0:
				print('Second')
			else:
				print('First')
		else:
			k = 0
			for i in range(n):
				if a[i] != 1:
					break
				k = i + 1
			if k % 2 == 0:
				print('First')
			else:
				print('Second')