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 collections import deque
from collections import Counter
from itertools import permutations
import heapq

def counterExample():
    arr = [1, 3, 4, 1, 2, 1, 1, 3, 4, 3, 5, 1, 2, 5, 3, 4, 5]
    counter = Counter(arr)
    top_three = counter.most_common(3)
    print(top_three)

q = deque()

#For stack we have:
#q.append()
#q.pop()

#For queue we have:
#q.append()
#q.popleft()


#for heap we have:
#heapify(a list)
#heapq.heappush()
#heapq.heappop()
#heapq.nlargest(k, iterable, key = fun)
#heapq.nsmallest(k, iterable, key = fun)



import math



def main():
    n = int(input())
    while n >= 1:
        n -= 1
        nr = int(input())
        l = [int(i) for i in input().split()]
        first = 1
        ok = 0
        for i in l:
            if i == 1:
                first = first ^ 1
            if i != 1:
                ok = 1
                if first == 1:
                    print("First")
                else:
                    print("Second")
                break
        if ok == 0:
            if first == 0:
                print("First")
            else:
                print("Second")








if __name__ == "__main__":
    main()