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.InputStreamReader;

public class B {
    public static void main(String[] args) {
        try {
            BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
            int t = Integer.parseInt(in.readLine());
            for (int i = 0; i < t; i++) {
                int n = Integer.parseInt(in.readLine());
                String[] s = in.readLine().split(" ");
                int[] a = new int[n];
                for (int j = 0; j < n; j++) {
                    a[j] = Integer.parseInt(s[j]);
                }
                int w = new B().solve(a, 0, 0);
                System.out.println(w == 0 ? "First" : "Second");
            }
            in.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    int solve(int[] a, int turn, int idx) {
        int ret = -1;
        if (idx == a.length) ret = 1 - turn;
        else {
            int v = a[idx];
            if (v == 1) {
                ret = solve(a, 1 - turn, idx + 1);
            } else {
                ret = turn;
            }
        }
        return ret;
    }
}