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. 
############## importing libraries ############################

from math import log2, ceil, floor
from collections import defaultdict, OrderedDict, Counter, deque
from heapq import heappush, heappop, heapify
import re

############### input functions ###############################
def inp():
    return input()
def list_inp():
    return list(map(int,input().split()))
def map_inp():
    return map(int,input().split())
def int_inp():
    return int(input())
def float_int():
    return float(input())
    
################# important functions ############################
def isprime(n):
    if (n <= 1) : 
        return False
    if (n <= 3) : 
        return True
    if (n % 2 == 0 or n % 3 == 0) : 
        return False
    i = 5
    while(i * i <= n) : 
        if (n % i == 0 or n % (i + 2) == 0) : 
            return False
        i = i + 6
    return True

################## my code ######################################

for _ in range(int_inp()):
    n = int_inp()
    l = list_inp()
    # l.reverse()
    turn = 1
    i=0
    # for i in range(n-1):
    #     if l[i]==1:
    #         if turn == 1:
    #             turn =2
    #         else:
    #             turn =1
    #     elif l[i]>1 and i==n-2:
    #         pass
    #     elif l[i]>1 and i!=n-2:
    #         if l[i+1]==1:
    #             if turn == 1:
    #                 turn =2
    #             else:
    #                 turn =1
    #         else:
    #             pass
    while(l[i]==1 and i<n):
        if turn == 1:
            turn =2
        else:
            turn =1
        i+=1
        if i>=n:
            break
        

    if i==n:
        if turn ==2:
            print("First")
        else:
            print("Second")
    else:
        if turn ==1:
            print("First")
        else:
            print("Second")