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. 
__author__ = 'Devesh Bajpai'

'''
https://codeforces.com/problemset/problem/1382/B

Solution: Till the players land on a pile having more than 1 stone in it, there isn't any strategy to employ. Each 
player picks the 1 stone in the pile. Once this ends, the player to play can strategically pick s-1 stones in the current
pile (having s stones), thereby forcing the next player to pick the one stone left. This makes the first player to be
in command again. If the player lands on the last pile, the strategy is to collect all the stones thereby
depriving the other player to play and losing. Hence we need to find the prefix of 1-stone piles in the given piles. 
If that count is even, the second player wins, otherwise the first player wins.
The exception to this strategy is the case where there are all 1s. As mentioned above, there isn't any strategy to 
employ and the number of the piles (n) determine who wins. If its odd, second player wins, otherwise the first player
wins.   
'''


def solve(n, arr):

    i = 0
    while i < n and arr[i] == 1:
        i += 1

    if i == n:
        if n % 2 == 0:
            return "Second"
        else:
            return "First"
    else:
        if i % 2 != 0:
            return "Second"
        else:
            return "First"


if __name__ == "__main__":

    t = int(raw_input())

    results = list()
    for _ in xrange(0, t):
        n = int(raw_input())
        arr = map(int, raw_input().split(" "))
        results.append(solve(n, arr))

    for result in results:
        print result
