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.io.*;
public class Solution{

    static class Pair<A, B> {

        A parent;
        B rank;

        Pair(A parent, B rank) {
            this.rank = rank;
            this.parent = parent;
        }
    }

    static class Node {
        int x;
        int y;
        Node(int k,int l)
        {
            x=k;y=l;
        }
    }
    static int m=1000000007;
    public static void main(String[] args) throws IOException {
        FastReader s1 = new FastReader();
        StringBuilder sb = new StringBuilder();
        int tt=s1.I();
        while(tt--!=0)
        {
            int n=s1.I();
            int ar[] = new int[n];
            for(int i=0;i<n;i++)
                ar[i]=s1.I();
            if(n==1)
            {
                sb.append("First\n");
                continue;
            }
            boolean ans[] = new boolean[n];
            ans[n-1]=true;
            for(int i=n-2;i>=0;i--)
            {
                if(ar[i]==1)
                {
                    ans[i]=(!ans[i+1]);
                }
                else
                {
                    ans[i]=((!ans[i+1])|ans[i+1]);
                }
            }
            if(ans[0])
            {
                sb.append("First\n");
            }
            else
                sb.append("Second\n");
        }
        System.out.println(sb);
    }
    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int I() {
            return Integer.parseInt(next());
        }

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

        double D() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static long gcd(long a, long b) {
        if (a % b == 0) {
            return b;
        }
        return gcd(b, a % b);
    }

    static float power(float x, int y) {
        float temp;
        if (y == 0) {
            return 1;
        }
        temp = power(x, y / 2);

        if (y % 2 == 0) {
            return temp * temp;
        } else {
            if (y > 0) {
                return x * temp * temp;
            } else {
                return (temp * temp) / x;
            }
        }
    }
    static long pow(long x, long y) {
        long res = 1;
        x = x % m;
        if (x < 0) {
            x += m;
        }
        while (y > 0) {
            if ((y & 1) == 1) {
                res = (res * x) % m;
                if (res < 0) {
                    res += m;
                }
            }
            y = y >> 1;
            x = (x * x) % m;
            if (x < 0) {
                x = x + m;
            }
        }
        res = res % m;
        if (res < 0) {
            res += m;
        }
        return res;
    }

    static void sieveOfEratosthenes(int n) {
        ArrayList<Integer> prime = new ArrayList<Integer>();
        boolean Prime[] = new boolean[n + 1];
        for (int i = 2; i < n; i++) {
            Prime[i] = true;
        }

        for (int p = 2; p * p <= n; p++) {
            if (Prime[p] == true) {
                prime.add(p);
                for (int i = p * p; i <= n; i += p) {
                    Prime[i] = false;
                }
            }
        }
    }
}