Problem: 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 (in JAVA):

Solution:
/*package com.company.tests.number658;*/

import com.sun.security.jgss.GSSUtil;

import java.util.*;

public class Main {
    public static Scanner key = new Scanner(System.in);

    //Common subsequence
    public static void solveA(int t) {
        while (t > 0) {
            t--;
            int n = key.nextInt();
            int m = key.nextInt();
            int[] a = new int[n];
            int[] b = new int[m];
            HashSet<Integer> set = new HashSet<Integer>();
            for (int i = 0; i < n; i++) {
                a[i] = key.nextInt();
                set.add(a[i]);
            }
            String ans = "NO";
            int common = -1;
            for (int i = 0; i < m; i++) {
                b[i] = key.nextInt();
                if(set.contains(b[i]) && common == -1) {
                    ans = "YES";
                    common = b[i];
                }
            }
            if(ans.equals("YES")) {
                System.out.println("YES");
                System.out.println("1 " + common);
            }
            else
                System.out.println("NO");
        }
    }



    public static boolean pilesRec(int[] a)
    {
        int count = 0;
        int countZ = 0;
        int x = -1;
        for (int i = 0; i < a.length; i++) {
            if(a[i] != 0 && x == -1)
                x = i;
            if(a[i] == 1)
                count++;
            if(a[i] == 0)
                countZ++;
        }
        if(x == -1)
            return false;
        if(count + countZ == a.length && count%2 != 0)
            return true;
        else if(a[x] > 1)
            return true;
        else
        {
            a[x] = 0;
            if(!pilesRec(a))
                return true;
        }
        return false;
    }


    //stones
    public static void solveB(int t)
    {
        while(t > 0){
            t--;
            int n = key.nextInt();
            int[] p = new int[n];
            for (int i = 0; i < n; i++) {
                p[i] = key.nextInt();
            }
            if(pilesRec(p))
                System.out.println("First");
            else
                System.out.println("Second");
        }
    }

    //prefix flip
    public static void solveC(int t)
    {
        while(t < 0)
        {
            t--;
            int l = key.nextInt();
            String a,b;
            a = key.next();
            b = key.next();
            char[] A = a.toCharArray();
            char[] B = b.toCharArray();
        }
    }


    public static void main(String[] args) {
        int t = key.nextInt();
        /*solveA(t);*/
        solveB(t);
    }
}