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 os
import random
import re
import sys
import copy
from collections import Counter
def bs(arr,x,l,r):
    if l<=r:
        mid=(r+l)//2
        if arr[mid]==x:
            return mid
        elif arr[mid]<x:
            return bs(arr,x,mid+1,r)
        else:
            return bs(arr,x,l,mid-1)
    else:
        return -1
def solve():
    for _ in range(int(input())):
        n=int(input())
        l=list(map(int,input().strip().split()))
        x=l.count(1)
        if n==1 or x==0:
            print("First")
            continue
        y=n-x
        i=0
        f=1
        while i<n:
            #print(i,l[i],x,y,f)
            if i==n-1:
                if f==1:
                    f=2
                else:
                    f=1
                break
            if l[i]==1:
                if f==1:
                    f=2
                else:
                    f=1
                x-=1
                i+=1
            elif l[i]>1:
                if l[i+1]==1:
                    j=i+1
                    c=0
                    while l[j]==1:
                        c+=1
                        j+=1
                        if j==n:
                            break
                    if j==n:
                        if c%2==0:
                            i+=1
                        else:
                            l[i]=1
                            x+=1
                    else: 
                        if c%2==0:
                            l[i]=1
                            x+=1
                        else:
                            i+=1
                else:
                    i+=1
                    y-=1
                    continue
                y-=1
                if f==1:
                    f=2
                else:
                    f=1
        if f==1:
            print("Second")
        else:
            print("First")
                    
            
if __name__=="__main__":
    solve()