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 generate_one_list(a):
    l = []; n = len(a); p = -1
    for i in range(n - 1, -1, -1):
        if a[i] == 1:
            l.append([-1, False])
        else:
            if p == -1:
                l.append([n - i - 1, True])
            else:
                l.append([p - i - 1, False])
            p = i
    return l[::-1]

def main():
    t = int(input())
    for _ in range(t):
        n = int(input())
        a = list(map(int, input().split()))
        turn = 0
        one_list = generate_one_list(a)
        z = 0
        while z < n - 1:
            if a[z] == 1:
                z += 1
            elif a[z + 1] == 1:
                one_train, r = one_list[z]
                if r:
                    if one_train % 2 != 0:
                        a[z] = 1
                    else:
                        z += 1
                else:
                    if one_train % 2 != 0:
                        z += 1
                    else:
                        a[z] = 1
            else:
                a[z] = 1
            turn += 1
        if turn % 2 == 0:
            print("First")
        else:
            print("Second")  # 7 1 1 1 is first
main()