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:
#for each pile except the last, a player can choose to complete it (take all 
#stones) or make the other person complete it (leave 1 stone)

from functools import lru_cache

@lru_cache(None)
def Iwillwin(pileNumber,nStones):
    if pileNumber==n-1: #I take everything in last pile
        return True
    else:
        if nStones>1 and Iwillwin(pileNumber,1)==False:
            return True #force enemy to lose by leaving 1 stone in current pile
        if Iwillwin(pileNumber+1,a[pileNumber+1])==False:
            return True #force opponent to lose by taking all stones in this pile and moving to next
        return False #no way to force opponent to lose

t=int(input())
for _ in range(t):
    n=int(input())
    a=[int(x) for x in input().split()]
    
    for i in range(n-2,-1,-1):
        Iwillwin(i,1) #to prevent recursion depth from exceeding

    if Iwillwin(0,a[0]):
        print('First')
    else:
        print('Second')
    Iwillwin.cache_clear()