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 java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

public class Two {

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        int t;
        t = in.nextInt();
        //t = 1;
        while (t > 0) {
            solver.call(in,out);
            t--;
        }
        out.close();
    }

    static class TaskA {
        public void call(InputReader in, PrintWriter out) {
            int n, a=0, b;
            n = in.nextInt();

            int[] arr = new int[n];

            for (int i = 0; i < n; i++) {
                arr[i] = in.nextInt();
            }

            for (int i = 0; i < n-1; i++) {
                if(i==0){
                    if(arr[i]==1) {
                        a++;
                        continue;
                    }
                    else
                        break;
                }
                if(arr[i]!=1)
                    break;
                a++;

            }

            if(a%2==0)
                b=1;
            else
                b=2;


            if(b==1){
                out.println("First");
            }
            else
                out.println("Second");

        }
    }

    static class answer implements Comparable<answer>{
        int a, b;

        public answer(int a, int b) {
            this.a = a;
            this.b = b;
        }


        @Override
        public int compareTo(answer o) {
            return this.a - o.a;
        }
    }


    static long lcm(long a , long b){
        return (a/gcd(a,b)) *b;
    }

    static long gcd(long a, long b)
    {
        if (b == 0)
            return a;
        return gcd(b,  a % b);
    }


    static final Random random=new Random();
    static void shuffleSort(int[] arr) {
        int n = arr.length;
        for (int i=0; i<n; i++) {

            int a=random.nextInt(n), temp=arr[a];
            arr[a]=arr[i];
            arr[i]=temp;

        }

        Arrays.sort(arr);

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong(){
            return Long.parseLong(next());
        }

    }
}