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.util.StringTokenizer;

public class Solution {
    private FastScanner in;
    private PrintWriter out;

    public static void main(String[] args) throws Exception {
        (new Solution()).run();
    }

    private void run() throws Exception {
        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    private void solve() throws Exception {
        int t = in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            int[] array = new int[n];
            for (int i = 0; i < n; i++) {
                array[i] = in.nextInt();
            }
            int i = 0;
            while (i < array.length && array[i] == 1) {
                i++;
            }
            out.println((i % 2 == 0) ^ (i == array.length)? "First" : "Second");
        }
    }

    private static class FastScanner implements Closeable {
        BufferedReader bufferedReader;
        StringTokenizer stringTokenizer;

        public FastScanner(InputStream inputStream) {
            this.bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
        }

        public FastScanner(File file) throws FileNotFoundException {
            this(new FileInputStream(file));
        }

        public String next() throws IOException {
            while (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
                String line = bufferedReader.readLine();
                if (line == null) {
                    throw new EOFException("End of input stream is reached.");
                }
                stringTokenizer = new StringTokenizer(line);
            }
            return stringTokenizer.nextToken();
        }

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

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

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        @Override
        public void close() throws IOException {
            bufferedReader.close();
        }
    }
}