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,stdout
from collections import Counter
from math import ceil
from bisect import bisect_left 
from bisect import bisect_right
import math

def ai(): return list(map(int, stdin.readline().split()))
def ei(): return map(int, stdin.readline().split())
def ip(): return  int(stdin.readline().strip())
def op(ans): return stdout.write(str(ans) + '\n')

for i in range(ip()):
    n = ip()
    piles = ai()
    w=i = 0
    while i<n-1 and piles[i]==1:
        w = ~w
        i += 1
    if w:print('Second')
    else:print('First')

        


# for i in range(ip()):
#     n,m = ei()
#     a = ai()
#     b = ai()
#     a = set(a)
#     b = set(b)
#     x = list(a.intersection(b))
#     if x:
#         print('YES')
#         print(1,x[0])
#     else:
#         print('NO')

#     n = ip()
#     s = input()
#     d = {}
#     for i in s:
#       if i not in d:
#           d[i] = 1
#       else:
#           d[i] += 1
#     ans = d.values()
#     cnt = 0
#     for i in ans:
#       if i%2:
#           cnt += 1
#           break
#     if cnt==0:
#       print("YES")
#     else:
#       print('NO')