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:
def cin():
    return map(int, input().split())


# for t in range(int(input())):
#     n, m = cin()
#     a = set(list(cin()))
#     b = list(cin())
#     ans = 'NO'
#     c = []
#     for i in b:
#         if i in a:
#             ans = 'YES'
#             c = [1, i]
#             break
#     print(ans)
#     if c:
#         print(*c)


for t in range(int(input())):
    n = int(input())
    a = list(cin())
    f = 1
    i = 0
    while i < n - 1 and a[i] == 1:
        f = (f + 1) % 2
        i += 1
    if f == 0:
        print('Second')
    else:
        print('First')
#     odd, even = 0, 0
#     curr = 0
#     i = 0
#     while i < n - 1:
#         j = i
#         while j < n - 1 and a[j] == 1:
#             j += 1
#         if j - i > 1:
#             odd += 1 if (j - i) % 2 else 0
#             even += 1 if not (j - i) % 2 else 0
#         i = j
#         while i < n and a[i] != 1:
#             i += 1
#     if not odd and not even:
#         print('First')
#     else:
#         print('oi')
    #
    # cnt1 = 0
    # for el in a:
    #     if el == 1:
    #         cnt1 += 1
    # if cnt1 == n - 1:
    #     if cnt1 % 2:
    #         print('Second')
    #     else:
    #         print('First')
    # else:
    #     if n == 1:
    #         print('First')


# for t in range(int(input())):
#     n = int(input())
#     a = input()
#     b = input()
#