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:

#Suppose a1>1. If removing the entire first pile is winning, player 1 will do that.
#Otherwise, player 1 can leave exactly one stone in the first pile, forcing player 2 to
#remove it, leaving player 1 in the winning position. Otherwise, if a1=1, then it is
#forced to remove the first pile.
#
#So, whichever player gets the first pile with more than one stone wins.
#That is, let k be the maximum number such that a1=⋯=ak=1. If k is even,
#the first player will win. Otherwise, the second player will win.
#The only exception is when all piles have exactly 1 stone.
#In that case, the first player wins when k is odd.


def c():
    n = int(input())
    flag = 0
    c = 0
    l = list(map(int, input().split()))
    if l[0] > 1:
        print("First")
        return
    elif l[0] == 1 and len(set(l)) == 1:
        if n % 2 == 0:
            print("Second")
            return
        else:
            print("First")
            return
    for i in l:
        if i == 1:
            c = c + 1
        else:
            break
    if c % 2 == 0:
        flag = 0
    else:
        flag = 1
    if flag == 0:
        print("First")
        return
    else:
        print("Second")
        return


for _ in range(int(input())):
   c()