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:


import java.math.*;
import java.util.*;
import java.io.*;
  
public class gr13b
{ 
    //By shwetank_verma
	static void addEdge(ArrayList<ArrayList<Integer>> l,int u,int v) {
		l.get(u).add(v);
		l.get(v).add(u);
	}
	
	public static void main(String[] args) 
    { 
        FastReader sc=new FastReader(); 
PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        
           
                int t=1;
                t=sc.nextInt();
                while(t-->0){
               int n=sc.nextInt();
               int a[]=new int[n];
               int no=0;
               for(int i=0;i<n;i++) {
            	   a[i]=sc.nextInt();
            	   
            	   
               }
               for(int i:a) {
            	   if(i==1)
            		   no++;
            	   else
            		   break;
               }
               if(no==n) {
            	   if(no%2!=0)
            		   pr.println("First");
            	   else
            		   pr.println("Second");
               }
               else if(no<n){
            	   if(no%2==0)
            		   pr.println("First");
            	   else
            		   pr.println("Second");
               }
               
                }
           
            
       
 pr.flush();
    }
    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 int mod=1000000007;
    static boolean primes[]=new boolean[1000007];
    static ArrayList<Integer> b=new ArrayList<>();
    static boolean seive(int n){
        Arrays.fill(primes,true);
        primes[0]=primes[1]=false;
        for(int i=2;i*i<=n;i++){
            if(primes[i]==true){
                for(int p=i*i;p<=n;p+=i){
                    primes[p]=false;
                }
            }
        }
        if(n<1000007){
        	for(int i=2;i<=n;i++) {
        		if(primes[i])
        			b.add(i);
        	}
            return primes[n];
        }
        return false;
        
    }
    static int gcd(int a,int b){
        if(b==0)
        return a;
        return gcd(b,a%b);
    }
    static long GCD(long a,long b){
        if(b==0)
        return a;
        return GCD(b,a%b);
    }
    static ArrayList<Integer> segseive(int l,int r){
    	
    	ArrayList<Integer> isprime=new ArrayList<Integer>();
    	boolean p[]=new boolean[r-l+1];
    	Arrays.fill(p, true);
        
    	for(int i=0;b.get(i)*b.get(i)<=r;i++) {
    		int currprime=b.get(i);
    		int base=(l/currprime)*currprime;
    		if(base<l) {
    			base+=currprime;
    		}
    		for(int j=base;j<=r;j+=currprime) {
    			p[j-l]=false;
    		}
    		
    		
    		if(base==currprime) {
    			p[base-l]=true;
    		}
    		
    	}
    	for(int i=0;i<=r-l;i++) {
    		if(p[i])
    			isprime.add(i+l);
    	}
    	return isprime;
    }
      static int LowerBound(int a[], int x) { // x is the target value or key
  int l=-1,r=a.length;
  while(l+1<r) {
    int m=(l+r)>>>1;
    if(a[m]>=x) r=m;
    else l=m;
  }
  return r;
}
static int UpperBound(int a[], int x) {// x is the key or target value
    int l=-1,r=a.length;
    while(l+1<r) {
       int m=(l+r)>>>1;
       if(a[m]<=x) l=m;
       else r=m;
    }
    return l+1;
 }

  
     
}