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.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class Sequential_Nim {
    static class RealScanner {
        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());
        }
    }

    public static void main(String[] args) {
        RealScanner sc = new RealScanner();
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            List<Integer> l = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                l.add(sc.nextInt());
            }
            List<String> s = new ArrayList<>();
            int count = 0;
            String[] str = {"First", "Second"};
            int k = 0;
            for (int i = n - 1; i >= 0; i--) {
                if (l.get(i) != 1) {
                    count = i + 1;
                    continue;
                }
                try {
                    if (l.get(i) == 1) {
                        s.add(str[k % 2]);
                      //  l.set(i, 0);
                        k++;
                    } else if (l.get(i + 1) == 1) {
                        s.add(str[k % 2]);
                      //  l.set(i, 0);
                        k++;
                    } else {
                        s.add(str[(k + 1) % 2]);
                       // l.set(i, 0);
                        k += 2;
                    }
                } catch (Exception e) {

                }
            }
          //  System.out.println(s);
            if (count == 0 && l.get(l.size() - 1) == 1) {
                if (s.get(s.size() - 1).equals("First")) {
                    System.out.println("First");
                } else {
                    System.out.println("Second");
                }
            } else if (count % 2 != 0) {
                System.out.println("First");
            } else {
                System.out.println("Second");
            }
//            if (s.size() == 0) {
//                System.out.println("First");
//            } else if (s.get(s.size() - 1).equals("Second")) {
//                System.out.println("First");
//            } else {
//                System.out.println("Second");
//            }
        }
    }
}