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:
import sys


def play(stones):
    stones_simple = []

    s = 0
    while s < len(stones):
        if s + 1 < len(stones) and stones[s] == 1 and stones[s + 1] == 1:
            s += 2
            continue

        stones_simple.append(stones[s])
        s += 1

    if not stones_simple:
        print("Second")
    elif len(stones_simple) == 1 and stones_simple[0] == 1:
        print("First")
    elif stones_simple[0] == 1:
        print("Second")
    else:
        print("First")


def play2(stones):
    ones = 0
    other = False
    for s in stones:
        if s == 1:
            ones += 1
        elif s != 1:
            other = True
            break

    if ones % 2 == 0 and other is False:
        print("Second")
    elif ones % 2 == 1 and other is False:
        print("First")
    elif ones % 2 == 0 and other is True:
        print("First")
    else:
        print("Second")


t = int(input())

i = 0
for line in sys.stdin:
    if i == 0:
        n = int(line)
    elif i == 1:
        stones = [int(x) for x in line.split()]

        play2(stones)

    i = 1 - i