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;
import java.util.Scanner;

public final class B {

    public static void main(String[] args) {
        final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        final int t = Integer.parseInt(in.nextLine());
        for (int x = 0; x < t; x++) {
            final int n = Integer.parseInt(in.nextLine());
            final int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = in.nextInt();
            }
            in.nextLine();
            System.out.println(dfs(arr, 0, new Boolean[n]) ? "First" : "Second");
        }
    }

    private static boolean dfs(int[] arr, int idx, Boolean[] dp) {
        if (idx == arr.length) {
            return false;
        }
        if (dp[idx] != null) {
            return dp[idx];
        }
        boolean res = false;
        final boolean next = !dfs(arr, idx + 1, dp);
        if (arr[idx] == 1) {
            res = next;
        } else {
            if (next) {
                res = true;
            } else {
                final int t = arr[idx];
                arr[idx] = 1;
                if (!dfs(arr, idx, dp)) {
                    res = true;
                }
                arr[idx] = t;
            }
        }
        return dp[idx] = res;
    }
}