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

public class B {
    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;
        public InputReader() throws FileNotFoundException {
            // reader = new BufferedReader(new FileReader("count.in"));
            reader = new BufferedReader (new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }
        String next() { // reads in the next String
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() { // reads in the next int
            return Integer.parseInt(next());
        }   
        public long nextLong() { // reads in the next long
            return Long.parseLong(next());
        }   
        public double nextDouble() { // reads in the next double
            return Double.parseDouble(next());
        }
    }

    public static void main (String[] args) throws IOException, FileNotFoundException {
	InputReader in = new InputReader();
        PrintWriter pw = new PrintWriter(System.out);
	int c = in.nextInt();
	for (int cases = 0; cases < c; cases++) {
	    int n = in.nextInt();
	    
	    // if (n == 1) { pw.println("First"); break; }
	    
	    int[] piles = new int[n];
	    boolean allOnes = true;
	    int inControl = 0;
	    boolean ICFound = false;
	    for (int i = 0; i < n; i++) {
	        piles[i] = in.nextInt();
	        if (piles[i] != 1) {
	            allOnes = false;
	            if (!ICFound) {
	                inControl = i + 1;
	                ICFound = true;
	            }
	        }
	    }
	    
	    if (allOnes) {
	       if (n % 2 == 0) pw.println("Second");
	       if (n % 2 == 1) pw.println("First");
	    }
	    // if (n == 1) pw.println("First");
	    else if (inControl % 2 == 0) pw.println("Second");
	    else pw.println("First");
	    
	}

	pw.close();
    }
}