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 sys import stdin
###############################################################
def iinput(): return int(stdin.readline())
def minput(): return map(int, stdin.readline().split())
def linput(): return list(map(int, stdin.readline().split()))
###############################################################

t = iinput()
while t:
    t-=1
    n = iinput()
    a = linput()
    temp = 1
    one = False if a[0] != 1 else True
    block = []
    for i in range(1, n):
        if (a[i] >1 and a[i-1]>1) or (a[i] == a[i-1] ==1):
            temp += 1
        else:
            block.append(temp)
            temp = 1
    block.append(temp)
    # print(block)
    d = {1: 'Second', 0: 'First'}
    beg = 0
    for e in range(len(block)-1):
        # print(beg)
        if one:
            if block[e] & 1:
                beg ^= 1
            one = False
        else:
            if e+1 == len(block)-1 and block[e+1]&1 == 0:
                beg ^= 1
            elif e+1 < len(block)-1 and block[e+1]&1:
                beg ^= 1
            else:
                pass
            one = True

    if one and block[-1]&1==0:
        beg ^= 1
    print(d[beg])