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.util.*;
import java.io.*;

public class Codeforces_1382_B {
	static FastReader sc;
	static PrintWriter pw;

	public static void main(String[] args) throws FileNotFoundException {
		boolean submit = true;
		sc = new FastReader(submit);
		pw = new PrintWriter(System.out);

		int T = sc.nextInt();
		for (int t = 0; t < T; t++) {

			solve();
			pw.flush();
		}

		pw.close();

	}

	static void solve() {
		// Solution
		String p1 = "First";
		String p2 = "Second";

		int n = _int();
		int[] Arr = _int(n);

		int lead = 0;
		for (int i : Arr) {
			if (i == 1) {
				lead++;
			} else {
				break;
			}
		}

		//Print(lead+"  ");
		if (lead == n) {
			if (n % 2 == 1)
				Println(p1);
			else
				Println(p2);
			return;
		}

		if (lead % 2 == 1) {
			Println(p2);
		} else {
			Println(p1);
		}

	}

	/*
	 * 
	 * 
	 * 
	 * 
	 * 
	 * 
	 * 
	 * 
	 * 
	 * 
	 * 
	 * 
	 * 
	 * 
	 * 
	 * 
	 * 
	 */

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader(boolean b) throws FileNotFoundException {
			if (b) {
				br = new BufferedReader(new InputStreamReader(System.in));
			} else {
				br = new BufferedReader(new FileReader("src/input.txt"));
			}

		}

		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 boolean isPrime(long n) {
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;

		if (n % 2 == 0 || n % 3 == 0)
			return false;

		for (long i = 5; i * i <= n; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;

		return true;
	}

	static void flush() {
		pw.flush();
	}

	static int _int() {
		return sc.nextInt();
	}

	static int[] _int(int n) {
		int[] Arr = new int[n];
		for (int i = 0; i < n; i++) {
			Arr[i] = sc.nextInt();
		}
		return Arr;
	}

	static int[][] _int(int m, int n) {
		int[][] Arr = new int[m][n];
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++) {
				Arr[i][j] = sc.nextInt();
			}
		}
		return Arr;
	}

	static long _long() {
		return sc.nextLong();
	}

	static long[] _long(int n) {
		long[] Arr = new long[n];
		for (int i = 0; i < n; i++) {
			Arr[i] = sc.nextLong();
		}
		return Arr;
	}

	static long[][] _long(int m, int n) {
		long[][] Arr = new long[m][n];
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++) {
				Arr[i][j] = sc.nextLong();
			}
		}
		return Arr;
	}

	static String _string() {
		return sc.next();
	}

	static String[] _string(int n) {
		String[] s = new String[n];
		for (int i = 0; i < n; i++) {
			s[i] = sc.next();
		}
		return s;
	}

	static char _char() {
		return sc.next().charAt(0);
	}

	static char[] _chars() {
		return sc.next().toCharArray();
	}

	static char[] _char(int n) {
		char[] c = new char[n];
		for (int i = 0; i < n; i++) {
			c[i] = sc.next().charAt(0);
		}
		return c;
	}

	static char[][] _chars(int n) {
		char[][] C = new char[n][];
		for (int i = 0; i < n; i++) {
			C[i] = _chars();
		}
		return C;
	}

	static char[][] _char(int m, int n) {
		char[][] C = new char[m][n];
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++) {
				C[m][n] = sc.next().charAt(0);
			}
		}
		return C;
	}

	static int max(int m, int n) {
		return m > n ? m : n;
	}

	static int min(int m, int n) {
		return m < n ? m : n;
	}

	static long max(long m, long n) {
		return m > n ? m : n;
	}

	static long min(long m, long n) {
		return m < n ? m : n;
	}

	static int max(Iterable<Integer> A) {
		int ans = Integer.MIN_VALUE;
		for (int i : A) {
			if (i > ans) {
				ans = i;
			}
		}
		return ans;
	}

	static int min(Iterable<Integer> A) {
		int ans = Integer.MAX_VALUE;
		for (int i : A) {
			if (i < ans) {
				ans = i;
			}
		}
		return ans;
	}

	static int max(int... A) {
		int ans = Integer.MIN_VALUE;
		for (int i : A) {
			if (i > ans) {
				ans = i;
			}
		}
		return ans;
	}

	static int min(int... A) {
		int ans = Integer.MAX_VALUE;
		for (int i : A) {
			if (i < ans) {
				ans = i;
			}
		}
		return ans;
	}

	static long max(long... A) {
		long ans = Long.MIN_VALUE;
		for (long i : A) {
			if (i > ans) {
				ans = i;
			}
		}
		return ans;
	}

	static long min(long... A) {
		long ans = Long.MAX_VALUE;
		for (long i : A) {
			if (i < ans) {
				ans = i;
			}
		}
		return ans;
	}

	static void Println() {
		pw.println();
	}

	static void Println(Object X) {
		pw.println(X.toString());
	}

	static void Print(long x) {
		pw.print(x);
	}

	static void Println(long x) {
		pw.println(x);
	}

	static void Print(String s) {
		pw.print(s);
	}

	static void Println(String s) {
		pw.println(s);
	}

	static void Println(String[] A) {
		for (String s : A) {
			pw.println(s);
		}
	}

	static void Print(String[] A) {
		for (String s : A) {
			pw.print(s + " ");
		}
		pw.println();
	}

	static void Print(int[] A) {
		for (int i : A) {
			pw.print(i + " ");
		}
		pw.println();
	}

	static void Println(int[] A) {
		for (int i : A) {
			pw.println(i);
		}
	}

	static void Println(int[][] A) {
		for (int[] a : A) {
			for (int i : a) {
				pw.print(i + " ");
			}
			pw.println();
		}
	}

	static void Sort(int[] A) {
		Arrays.sort(A);
	}

	static void Sort(List<Integer> A) {
		Collections.sort(A);
	}

	static void Sortr(int[] A) {
		Arrays.sort(A);
		reverse(A);
	}

	static void reverse(int[] A) {
		int n = A.length;
		if (n < 2)
			return;
		int temp;
		for (int i = 0; i < n / 2; i++) {
			temp = A[i];
			A[i] = A[n - i - 1];
			A[n - i - 1] = temp;
		}
	}

	static void Sortr(List<Integer> A) {
		Collections.sort(A, Collections.reverseOrder());
	}

}
