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.IOException;
import java.io.InputStream;
import java.util.ArrayList;


public class Solution {
    public static void main(String[] args) throws IOException {
        int i, j;
        FastReader in = new FastReader(System.in);
        StringBuilder sb = new StringBuilder();
        boolean testcases = true;
        int t = testcases ? in.nextInt() : 1;
        int to = 0;
        while (to++ < t) {
            int n=in.nextInt();
            int arr[]=new int[n+1];
            boolean win[]=new boolean[n+1];
            for(i=1;i<=n;i++)
                arr[i]=in.nextInt();
            win[n]=true;
            for(i=n-1;i>=1;i--){
                if(win[i+1]){
                    if(arr[i]>1)
                        win[i]=true;
                }
                else{
                    win[i]=true;
                }
            }
            if(win[1])
                sb.append("First\n");
            else
                sb.append("Second\n");
        }
        System.out.print(sb);
    }
}
class FastReader {

    byte[] buf = new byte[2048];
    int index, total;
    InputStream in;

    FastReader(InputStream is) {
        in = is;
    }

    int scan() throws IOException {
        if (index >= total) {
            index = 0;
            total = in.read(buf);
            if (total <= 0) {
                return -1;
            }
        }
        return buf[index++];
    }

    String next() throws IOException {
        int c;
        for (c = scan(); c <= 32; c = scan()) ;
        StringBuilder sb = new StringBuilder();
        for (; c > 32; c = scan()) {
            sb.append((char) c);
        }
        return sb.toString();
    }

    int nextInt() throws IOException {
        int c, val = 0;
        for (c = scan(); c <= 32; c = scan()) ;
        boolean neg = c == '-';
        if (c == '-' || c == '+') {
            c = scan();
        }
        for (; c >= '0' && c <= '9'; c = scan()) {
            val = (val << 3) + (val << 1) + (c & 15);
        }
        return neg ? -val : val;
    }

    long nextLong() throws IOException {
        int c;
        long val = 0;
        for (c = scan(); c <= 32; c = scan()) ;
        boolean neg = c == '-';
        if (c == '-' || c == '+') {
            c = scan();
        }
        for (; c >= '0' && c <= '9'; c = scan()) {
            val = (val << 3) + (val << 1) + (c & 15);
        }
        return neg ? -val : val;
    }
}


