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. 
import com.sun.jdi.ArrayReference;
import com.sun.source.tree.WhileLoopTree;

import java.io.*;
import java.util.*;
import java.util.concurrent.atomic.AtomicReferenceFieldUpdater;

public class CP {
    static long startTime;
    static long endTime;
    static Boolean [] prime ;
    static Scanner sc = new Scanner(System.in);
    static class FastReader{
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static int lowerBound
            (Integer[] array, int length, int value) {
        int l = 0;
        int r = length;
        while (l < r) {
            int mid = (l + r) / 2;
            if (value < array[mid]) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
    public static long gcd(long a, long b){
        if (b == 0)
            return a;
        else return gcd(b, a % b);
    }
    public  static  long countDivs(long n ){
        int cn=0;
        long sqr = (long)Math.sqrt(n);
        for (long i = 1; i<=sqr ; ++i) {
            if(n%i==0){
                ++cn;
            }
        }
        cn*=2;
        if(sqr*sqr==n) cn--;
        return cn;

    }
    static void prime(int  x) {
        //sieve algorithm. nlog(log(n)).
        prime=new Boolean[ (x+1)];
        Arrays.fill(prime,true);
        prime[0] = prime[1] = false;
        for (int  i = 2; i * i <= x ; i++)
            if (prime[i])
                for (int  j = i * i; j <= x; j += i)
                    prime[j] = false;
    }
    static int[] sum;
    static void cumulitiveSum( int [] arr){
        sum = new int[arr.length];
        sum[0]=arr[0];
        for (int i = 1; i <arr.length; i++) {
            sum[i]=arr[i]+sum[i-1];
        }
    }
    static boolean isEven(long x ){
        return x % 2 == 0;
    }
    static boolean isPrime(long x ){
        boolean flag = true;
        int sqr = (int)Math.sqrt(x);
        if(x<2) return false;
        for (int i = 2; i <=sqr; i++) {
            if(x%i==0){
                flag=false;
                break;
            }
        }
        return flag;
    }
    public static void main(String [] args)   {
        startTime = System.currentTimeMillis();
        int T = sc.nextInt();while (T--!=0)
        {
            solve();
        }
        endTime = System.currentTimeMillis();
        long duration= endTime-startTime;
        //System.out.println(duration);
    }
    public static void solve()  {
//////////////////////////////////////////////////////////////////////
        int n =sc.nextInt();
        int [] arr=  new int[n];String f="First",s="Second";
        int one=0;
        for (int i = 0; i < n; i++) {
            arr[i]=sc.nextInt();
            if(arr[i]==1) one++;
        }
        if(one==n){
            if(n%2==0){
                System.out.println(s);

            }
            else {
                System.out.println(f);
            }
            return;
        }
        int preone=0;
        for (int i = 0; i < n; i++) {
            if(arr[i]==1) preone++;
            else break;
        }

        if(preone%2==0){
            System.out.println(f);
        }
        else {
            System.out.println(s);

        }


///////////////////////////////////////////////////////////////////////
    }

}