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.*;
import java.util.*;

public class A
{
    static int n;
    static int[] arr;
    static char[] s;
    public static void main(String[] args) throws IOException
    {
        Flash f = new Flash();
        int T = f.ni();
        for(int tc = 1; tc <= T; tc++){
            n = f.ni(); arr = f.arr(n); sop(fn());
        }
    }
    
    static String fn()
    {
        int i;
        for(i = 0; i < n; i++) if(arr[i] != 1) break;
        
        if(i == n && i % 2 == 0) return "Second";
        if(i == n && i % 2 == 1) return "First";
        
        if(i % 2 == 0) return "First";
        return "Second";
    }
    
    static void sort(int[] a){
        List<Integer> A = new ArrayList<>();
        for(int i : a) A.add(i);
        Collections.sort(A);
        for(int i = 0; i < A.size(); i++) a[i] = A.get(i);
    }
    
    static int swap(int itself, int dummy){
        return itself;
    }
    
    static void sop(Object o){ System.out.println(o); }
    
    static void print(int[] a){
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < a.length; i++) sb.append(a[i] + " ");
        System.out.println(sb);
    }
    
    static class Flash
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        
        String next(){
            while(!st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch(IOException e){
                    e.printStackTrace();
                }
            }
            
            return st.nextToken();
        }
        
        String ns(){
            String s = new String();
            try{
                s = br.readLine().trim();
            }catch(IOException e){
                e.printStackTrace();
            }
            
            return s;
        }
        
        int ni(){
            return Integer.parseInt(next());
        }
        
        int[] arr(int n){
            int[] a = new int[n];
            for(int i = 0; i < n; i++) a[i] = ni();
            return a;
        }

        long nl(){
            return Long.parseLong(next());
        }
        
        double nd(){
            return Double.parseDouble(next());
        }
    }
}