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.*;
import java.math.BigDecimal;
import java.math.MathContext;
import java.util.*;


/*
 * @author: Ramanqul
 * */
public class Main {
    private static boolean USE_TEST_CASES = true;
    private static final double EPS = .000000001;

    private static boolean contains(String s, String sub, int start) {
        for (int i=0;i<sub.length();i++) {
            if (start+i >= s.length()) {
                return false;
            }

            char c = s.charAt(start+i);
            if (c != '?' && c != sub.charAt(i)) {
                return false;
            }
        }

        return true;
    }

    private static int countOccurences(String s, String required) {
        int occurences = 0;
        int prev = -1;

        while (true) {
            prev = s.indexOf(required, prev+1);
            if (prev < 0) {
                break;
            }
            occurences++;
        }

        return occurences;
    }

    private static void solve(final int testCase, MyScanner in, PrintWriter out) {
        int n = in.nextInt();
        int a[] = new int[n];

        H.readArray(in, a, n);

        int cnt = 0;

        for (int i=0;i<n-1;i++) {
            if (a[i] == 1) {
                cnt++;
            } else {
                break;
            }
        }

        if (n == 1 || cnt % 2 == 0) {
            out.println("First");
        } else {
            out.println("Second");
        }

    }



    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int t = 1;

        if (USE_TEST_CASES) {
            t = sc.nextInt();
        }

        for (int i=0;i<t;i++) {
            solve(i, sc, out);
        }
        out.close();
    }

    private static class Pair<A,B> {
        public A first;
        public B second;
        public Pair(A first, B second) {
            this.first = first;
            this.second = second;
        }
    }

    private static class Triple<A,B,C> {
        public A first;
        public B second;
        public C third;

        public Triple(A first, B second, C third) {
            this.first = first;
            this.second = second;
            this.third = third;
        }
    }

    private static class H {
        public static void readArray(MyScanner in, int a[], int n) {
            for (int i=0;i<n;i++) {
                a[i] = in.nextInt();
            }
        }

        public static boolean isUnique(int ... nums) {
            Set<Integer> s = new HashSet<>();
            for (int i:nums) {
                if (s.contains(i)) {
                    return false;
                }
                s.add(i);
            }

            return true;
        }


        public static Pair<Integer, Integer>[] readIntPairs(MyScanner in, int n) {
            Pair[] pairs = new Pair[n];
            for (int i=0;i<n;i++) {
                int x = in.nextInt();
                int y = in.nextInt();
                pairs[i] = new Pair<Integer, Integer>(x, y);
            }

            return pairs;
        }

        public static List<Integer> dfs(int start, List<Integer>[] adj) {
            boolean v[] = new boolean[adj.length+1];

            List<Integer> res = new ArrayList<>();

            Stack<Integer> st = new Stack<>();

            st.push(start);

            while (!st.isEmpty()) {
                int x = st.pop();
                if (!v[x]) {
                    res.add(x);
                    boolean allVisited = true;
                    for (int e: adj[x]) {
                        if (!v[e]) {
                            st.push(e);
                            allVisited = false;
                        }
                    }

                    if (allVisited) {
                        break;
                    }

                    v[x] = true;
                } else {
                    break;
                }
            }

            return res;
        }

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

            return gcd(b%a, a);
        }

        public static long lcm(long a, long b) {
            return (a * b) / gcd(a,b);
        }
    }

    private static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        public MyScanner(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

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

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

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

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

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

    }
}