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.math.*;
import java.util.*;

public class Sample
{

	static int[][][] dp;
	static int[] are;

	public static void main(String[] args) throws InterruptedException {
		Thread t=new Thread(null, null,"", 1<<28) {
			public void run() {
				m();
			}
		};
		t.start();
		t.join();
	}

	public static void m()
	{
		FastScanner in=new FastScanner();
		int t = in.nextInt();
		while(t-->0)
		{
			int n = in.nextInt();
			int[] are = in.readArray(n);
			int change=0;
			for(int i=0; i<n; i++)
			{
				if(i==n-1) break;
				if(are[i]!=1) break;
				change++;
			}
			System.out.println(change%2==1 ? "Second" : "First");
		}
	}

	static void longestsub(int[] str1, int[] str2)
	{
		int n = str1.length;
		int m = str2.length;
		//
		int[][] dp = new int[n+1][m+1];
		//
		for(int i=0; i<=n; i++)
		{
			for(int j=0; j<=m; j++)
			{
				if(i==0 || j==0) {dp[i][j]=0; continue;}
				//
				if(str1[i-1]==str2[j-1]) dp[i][j]=dp[i-1][j-1]+1;
				else dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
			}
		}
		//
		int length = dp[n][m];
		//
		String ans = "";
		//
		int i=n, j=m;
		while(true)
		{
			if(i==0 || j==0) break;
			//
			if(dp[i][j]==Math.max(dp[i-1][j], dp[i][j-1]))
			{
				if(dp[i-1][j]>dp[i][j-1]) i--;
				else j--;
			}
			else
			{
				ans = str1[i-1] + " " + ans;
				i--; j--;
			}
		}
		//
		System.out.println(length>0?"YES"+'\n'+"1 "+ans.substring(0,1):"NO");
	}

	static void print(String[] are)
	{
		for(int i=0; i<are.length; i++)
			System.out.print(are[i]);
		System.out.println();
	}

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

	static class FastScanner
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");

		public String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();

		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}

}
