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.lang.*;
import java.io.*;

public class Three_Pairwise_Maximums{
  public static void main(String[] args) throws IOException{
    StringBuilder print = new StringBuilder();
    BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
    long t = Long.parseLong(input.readLine());
    while(t-- > 0){

      int n = Integer.parseInt(input.readLine());
      StringTokenizer st = new StringTokenizer(input.readLine());
      int binaryOutcome = 1;
      boolean first1s = true;
      for(int i = 0; i < n; i++){
        int a = Integer.parseInt(st.nextToken());
        if(first1s){
          if(a == 1){
            if(i != n-1){
              binaryOutcome += (1 - binaryOutcome) + (2 - binaryOutcome);
            }
          }else{
            first1s = false;
          }
        }


      }
      if(binaryOutcome == 1){
        print.append("First\n");
      }else{
        print.append("Second\n");
      }
    }
    System.out.println(print.toString());
  }
}