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.util.*;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.io.*;
import java.math.*;
import static java.lang.Math.min;
import static java.lang.Math.max;
import static java.lang.Math.abs;

/* docstring*/
public class c1324C {
    static FastScanner fs = new FastScanner();
    static PrintWriter fop = new PrintWriter(System.out);
    static StringBuilder sb = new StringBuilder();

    public static void main(String[] args) {
        try {
            solve();
        } catch (Exception e) {
            return;
        }
    }

    static void solve() throws IOException {

        int T = fs.nextInt();
        while (T-- > 0) {
            // write your code her
            int n=fs.nextInt();
            int[] a=fs.readArray(n);
            boolean dp[]  = new boolean[n+1];
            dp[n-1]=true;
            for(int i=n-2;i>=0;i--){
                if(dp[i+1]==false){
                    dp[i]=true;
                }
                else if (a[i]>1){
                    dp[i]=true;
                }
            }
            fop.println(dp[0]?"First":"Second");
        }
        fop.flush();
        fop.close();
    }

    static int count(int h) {
        return h * (3 * h + 1) / 2;
    }

    static class FastScanner {
        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[] readLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

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