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
from math import floor as fl
from math import ceil as ce
from math import sqrt
cont = sys.stdin.readlines()
cont = [i.strip('\n').split() for i in cont]
num = int(cont.pop(0)[0])

def arrtoint(arr):
    return [int(i) for i in arr]

for i in range(num):
    arr = cont[:2]
    cont = cont[2:]
    l = int(arr.pop(0)[0])
    arr = arrtoint(arr[0])

    #edge cases
    if len(arr) == 1:
        print("First")
    elif arr.count(1) == l:
        if l % 2 == 1:
            print("First")
        else:
            print("Second")
    elif arr.count(1) == 0:
        print("First")
    else:
        #edge cases are done

        arr.pop()

        if set(arr) == {1}:
            if l % 2 == 1:
                print("First")
            else:
                print("Second")
            continue
        #if the number of leading ones is eve, first
        #else, second

        aind = arr.index(next(filter(lambda x: x>1, arr)))
        if aind % 2 == 0:
            print("First")
        else:
            print("Second")