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:
# t = int(input())
# def count(i,vt,arr):
#     phu = 0
#     for y in range(i, vt+1):
#         if arr[y] > 1:
#             phu = phu + 2
#         else:
#             phu = phu + 1
#     return phu
# while t:
#     n = int(input())
#     l = []
#     l1 = []
#     l = input().split()
#     for _ in l:
#         l1.append(int(_))
#     ms = 1
#     for i in range(len(l1)-2,-1,-1):
#         if l1[i] == 1:
#             ms += 1
#             if ms % 2 == 0:
#                 if i <= 2:
#                     vt = i
#                     break
#         else:
#             vt = i
#             break
#     winner = 1
#     i=0
#     while i != vt+1:
#         if l1[i] == 1:
#             i += 1
#             if winner == 1:
#                 winner = 2
#             else:
#                 winner = 1
#         else:
#             able = count(i, vt, l1)
#             if able > 2:
#                 if able % 2 == 0:
#                     if winner == 1:
#                         winner = 2
#                     else:
#                         winner = 1
#                     l1[i] = 1
#                 else:
#                     if winner == 1:
#                         winner = 2
#                     else:
#                         winner = 1
#                     i += 1
#             else:
#                 if ms % 2 == 0:
#                     if winner == 1:
#                         winner = 2
#                     else:
#                         winner = 1
#                     i += 1
#                 else:
#                     l1[i] = 1
#                     if winner == 1:
#                         winner = 2
#                     else:
#                         winner = 1
#     print(winner)

t = int(input())
while t:
    n = int(input())
    l1 = list(map(int, input().split()))

    win = True
    for i in range(len(l1)):
        if l1[i] != 1:
            break
        if i + 1 != len(l1):
            win = not win
        else:
            break
    if win:
        print('First')
    else:
        print('Second')
    t -= 1