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 javax.print.DocFlavor;
import javax.print.attribute.HashAttributeSet;
import java.awt.event.MouseAdapter;
import java.io.*;

import java.math.BigInteger;
import java.nio.channels.FileChannel;
import java.util.*;

import java.util.ArrayList;

import java.util.Scanner;


public class adaking {

    static class pair {
        int first;
        int second;

        pair(int a, int b) {
            first = a;
            second = b;
        }
    }

    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;
        }
    }

    static final int mod = 1000000007;

    public static int gcd(int a, int b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
    public static String findSubString(String str, String pat)
    {
        int len1 = str.length();
        int len2 = pat.length();

        // check if string's length is less than pattern's
        // length. If yes then no such window can exist
        if (len1 < len2)
        {
            System.out.println("No such window exists");
            return "";
        }

        int hash_pat[] = new int[256];
        int hash_str[] = new int[256];

        // store occurrence ofs characters of pattern
        for (int i = 0; i < len2; i++)
            hash_pat[pat.charAt(i)]++;

        int start = 0, start_index = -1, min_len = Integer.MAX_VALUE;

        // start traversing the string
        int count = 0; // count of characters
        for (int j = 0; j < len1 ; j++)
        {
            // count occurrence of characters of string
            hash_str[str.charAt(j)]++;

            // If string's char matches with pattern's char
            // then increment count
            if (hash_pat[str.charAt(j)] != 0 &&
                    hash_str[str.charAt(j)] <= hash_pat[str.charAt(j)] )
                count++;

            // if all the characters are matched
            if (count == len2)
            {
                // Try to minimize the window i.e., check if
                // any character is occurring more no. of times
                // than its occurrence in pattern, if yes
                // then remove it from starting and also remove
                // the useless characters.
                while ( hash_str[str.charAt(start)] > hash_pat[str.charAt(start)]
                        || hash_pat[str.charAt(start)] == 0)
                {

                    if (hash_str[str.charAt(start)] > hash_pat[str.charAt(start)])
                        hash_str[str.charAt(start)]--;
                    start++;
                }

                // update window size
                int len_window = j - start + 1;
                if (min_len > len_window)
                {
                    min_len = len_window;
                    start_index = start;
                }
            }
        }

        // If no window found
        if (start_index == -1)
        {
            System.out.println("No such window exists");
            return "";
        }

        // Return substring starting from start_index
        // and length min_len
        return str.substring(start_index, start_index + min_len);
    }
    public static void primeFactors(long n)
    {
        // Print the number of 2s that divide n
        long ans=0;
        long m=n;
        while (n%2==0)
        {
            //System.out.print(2 + " ");
            ans=2;
            n /= 2;
        }

        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i+= 2)
        {
            // While i divides n, print i and divide n
            while (n%i == 0)
            {
               // System.out.print(i + " ");
                ans=i;
                n /= i;
            }
        }

        // This condition is to handle the case whien
        // n is a prime number greater than 2
        //System.out.println(ans);
        if (n > 2)
        {
            //System.out.print(n);
            ans=n;

        }
        if(ans==m)
            System.out.println(1+" "+(ans-1));
        else
        System.out.println(ans+" "+(m-ans));

    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        for (int k = 0; k < t; k++)
        {
            int n=sc.nextInt();

            int a[]=new int[n];
            for(int j=0;j<n;j++)
            {
                a[j]=sc.nextInt();
            }


            int f=0;
            for(int j=0;j<n;j++)
            {
                //System.out.println(a[j]);
                if(a[j]==1)
                    f++;
                else
                    break;
            }
            //System.out.println(f);
            if(f==0)
                System.out.println("First");
            else if(f==n)
            {
                if(f%2==0)
                    System.out.println("Second");
                else
                    System.out.println("First");
            }
            else if(f%2==0)
                System.out.println("First");
            else
                System.out.println("Second");






        }
    }
}




















