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. 
from __future__ import division, print_function

import sys
from atexit import register

if sys.version_info[0] < 3:
    from io import BytesIO as stream
else:
    from io import StringIO as stream


sys.stdin = stream(sys.stdin.read())
input = lambda: sys.stdin.readline().rstrip('\r\n')

sys.stdout = stream()
register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))


def read_int():
    return int(input())


def read_ints():
    return list(map(int, input().split(' ')))


def solve():
    """
    5 5 1

    1 x x x - second
    x 1 1 x 
    
    1 4
    dp[n] - can reach n-th on my turn
    dp[n] = a[n-1] > 1 and dp[n-1] or a[n-1] == 1 and dp[n-2]
    dp[1] = True
    dp[2] = 
    """
    n = read_int()
    A = read_ints()
    dp = [True]
    for i in range(1, n):
        if i > 1:
            dp.append((A[i-1] > 1 and dp[i-1]) or (A[i-1] == 1 and dp[i-2]))
        else:
            dp.append(A[i-1] > 1 and dp[i-1])
    if dp[-1]:
        return 'First'
    return 'Second'


if __name__ == '__main__':
    T = read_int()
    for _ in range(T):
        print(solve())
