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

def gcd(a,b): 
    if (b == 0): 
         return a 
    return gcd(b, a%b) 

def lcm(a,b): 
    return (a*b) / gcd(a,b) 

def bs(arr, l, r, x): 
    while l <= r: 
        mid = l + (r - l)//2; 
        if(arr[mid]==x): 
            return arr[mid]
        elif(arr[mid]<x): 
            l = mid + 1
        else: 
            r = mid - 1
    return -1

def swap(list, pos1, pos2): 
      
    list[pos1], list[pos2] = list[pos2], list[pos1] 
    return list

def subarrayBitwiseOR(A): 
    res = set()
    pre = {0} 
    for x in A: 
        pre = {x | y for y in pre} | {x} 
        res |= pre 
    return len(res) 

t = int(input())
for _ in range(t):
    n = int(input())
    i = 0
    a = list(map(int,input().split()))
    f=0
    while(i<n):
        if(a[i]>1):
            if(i%2==0):
                print("First")
            else:
                print("Second")
            f = 1
            break
        i+=1
    if(f):
        continue
    if(n%2!=0):
        print("First")
    else:
        print("Second")