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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main {
static long merge(long a[],long s,long e){
    long mid=(s+e)/2;
    long i=s;
    long j=mid+1;
    long k=s;
    long temp[]= new long[a.length];
    long count=0;
    while (i<=mid&&j<=e){
        if (a[(int)i]<=a[(int)j]){
            temp[(int)k]=a[(int)i];
            i++;
            k++;
        }
        else{
            temp[(int)k++]=a[(int)j++];
            count+=mid-i+1;
        }
    }
    while (i<=mid){
        temp[(int)k++]=a[(int)i++];
    }
    while (j<=e){
        temp[(int)k++]=a[(int)j++];
    }
    for (i=s;i<=e;i++){
        a[(int)i]=temp[(int)i];
    }
    return count;
}
static long find(long a[],long s,long e){
    if (s>=e)return 0;
    long mid=(s+e)/2;
    long x=find(a,s,mid);
    long y=find(a,mid+1,e);
    long z=merge(a,s,e);
    return x+y+z;
}
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- >= 1) {
            long n = sc.nextLong();
            long a[] = new long[(int) n];
            for (long i = 0; i < n; i++) {
                a[(int) i] = sc.nextLong();
            }
            boolean ans=true;
            for (int i=0;i<n;i++){
                if (a[i]==1){
                    ans=!ans;
                }
                else{
                    ans=!ans;
                    break;
                }
            }
            if (ans==false){
                System.out.println("First");
            }
            else System.out.println("Second");

        }
    }
    static void sort(int[] a) {
        ArrayList<Integer> l=new ArrayList<>();
        for (int i:a) l.add(i);
        Collections.sort(l);
        for (int i=0; i<a.length; i++) a[i]=l.get(i);
    }

    static class FastScanner {
        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();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
         int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }


}