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:
# cook your dish here
#code
#    ___________________________________
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#   |        _,     _   _     ,_        |
#   |    .-'` /     \'-'/     \ `'-.    |
#   |   /    |      |   |      |    \   |
#   |  ;      \_  _/     \_  _/      ;  |
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#   |  ;    .-.   .-.   .-.   .-.    ;  |
#   |   \  (   '.'   \ /   '.'   )  /   |
#   |    '-.;         V         ;.-'    |
#   |        `                 `        |
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#   |___________________________________|
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#   |  Author      :   Ramzz            |
#   |  Created On  :   21-07-2020       |
#   |___________________________________|
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#   | '__/ _` | '_ ` _ \|_  /_  /
#   | | | (_| | | | | | |/ / / / 
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#

import math
import collections
from sys import stdin,stdout,setrecursionlimit
from bisect import bisect_left as bsl
from bisect import bisect_right as bsr
import heapq as hq
setrecursionlimit(2**20)

t = 1
t = int(stdin.readline())

for _ in range(t):
    n = int(stdin.readline())
    #s = str(stdin.readline().strip('\n'))
    a = list(map(int, stdin.readline().rstrip().split()))
    
    if(n==1):
        print('First')
        continue
    else:
        s = sum(a)
        if(s==n and n%2==0):
            print('Second')
        elif(s==n and n%2==1):
            print('First')
        else:
            ans = 0
            for i in a:
                ans+=1
                if(i!=1):
                    break
            if(ans%2==1):
                print('First')
            else:
                print('Second')