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 for: http://codeforces.com/contest/1382/problem/B */

import java.util.*;
import java.io.*;
import java.util.Arrays; 
import java.util.ArrayList; 
import java.util.Collections; 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.StringTokenizer;

public class Solution {


    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        FastScanner in = new FastScanner(System.in);
        String scanInput;

        int loop = in.nextInt();

        for (int z = 0; z < loop; z++) {
            int n = in.nextInt();
            ArrayList <Integer> a = in.nextIntList(n);
            ArrayList <Character> c = new ArrayList<>();

            for (int i = n-1; i>=0; i--) {
                if (c.size() == 0) {
                    c.add('x');
                } else {
                    if(c.get(c.size()-1) == 'x') {
                        if (a.get(i) == 1) {
                            c.add('y');
                        } else {
                            c.add('y');
                            c.add('x');
                        } 
                    } else {
                        c.add('x');
                    }
                }  
            }
            //out.println(c);
            if(c.get(c.size()-1) == 'x') {
                out.println("First");
            } else {
                out.println("Second");
            }

        }
        out.close();
    }
}



class FastScanner {
   private final BufferedReader br;
   private StringTokenizer st;
 
    FastScanner(InputStream InputStream) {
        br = new BufferedReader(new InputStreamReader(InputStream));
    }
    
    String next() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }
    
    int nextInt() {
        return Integer.parseInt(next());
    }
    
    long nextLong() {
        return Long.parseLong(next());
    }
    
    double nextDouble() {
        return Double.parseDouble(next());
    }


    
    ArrayList<Integer> nextIntList(int count) {
        ArrayList<Integer> array = new ArrayList<>();
        for (int n = 0; n < count; n++) {
            int number = nextInt();
            array.add(number);
        }
        return array;
    }
    
    int[] nextIntArray(int count) {
        int[] array = new int[count];
        for (int n = 0; n < count; n++) {
            array[n] = nextInt();
        }
        return array;
    }
}



