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
from bisect import bisect_left as bl
from bisect import bisect_right as br

def input():
    return stdin.readline()[:-1]


def intput():
    return int(input())


def sinput():
    return input().split()


def intsput():
    return map(int, sinput())


class RangedList:
    def __init__(self, start, stop, val=0):
        self.shift = 0 - start
        self.start = start
        self.stop = stop
        self.list = [val] * (stop - start)

    def __setitem__(self, key, value):
        self.list[key + self.shift] = value

    def __getitem__(self, key):
        return self.list[key + self.shift]

    def __repr__(self):
        return str(self.list)

    def __iter__(self):
        return iter(self.list)


def dprint(*args, **kwargs):
    if debugging:
        print(*args, **kwargs)


debugging = 1
# Code
t = intput()
for _ in range(t):
    n = intput()
    a = list(intsput())
    cnt = 0
    for x in a:
        if x == 1:
            cnt += 1
        else:
            break
    if cnt == n:
        if cnt % 2:
            print('First')
        else:
            print('Second')
    elif cnt % 2:
        print('Second')
    else:
        print('First')