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 math
import sys
from sys import stdin
input=sys.stdin.readline
def solve(p, n): 
    G = 0
    for i in range(n):  
        if (p[i] % 2 != 0):  
            G ^= (p[i] + 1)  
        else:  
            G ^= (p[i] - 1)  
    return G 
for _ in range(int(input())):
    n=int(input())
    arr=[int(j) for j in input().split()]
    if n==2 and arr[0]==1:
        print("Second")
    elif n==1:
        print("First")
    else:
        count1 = arr.count(1)
        dic=dict()
        for i in arr:
            if i not in dic:
                dic[i]=1
            else:
                dic[i]+=1
        if(count1 == n):
            if(n%2==1):
                print("First")
            else:
                print("Second")
        elif(count1 == 0):
            print("First")
        else:
            co = 0
            for i in range(n):
                if(arr[i]==1):
                    co += 1
                else:
                    break
            if(co%2!=0):
                print("Second")
            else:
                print("First")