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.io.*;
import java.math.BigInteger;
import java.util.*;
public class Codeforce {

    static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Reader(String file_name) throws IOException
        {
            din = new DataInputStream(
                    new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String readLine() throws IOException
        {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') {
                    if (cnt != 0) {
                        break;
                    }
                    else {
                        continue;
                    }
                }
                buf[cnt++] = (byte)c;
            }
            return new String(buf, 0, cnt);
        }

        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ') {
                c = read();
            }
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0,
                    BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }
    static int mod = (int) (1e9 + 7);

    public static class pair implements Comparator<pair> {
        int x;
        int y;

        public pair() {

        }

        public pair(int x, int y) {
            this.x = x;
            this.y = y;

        }

        @Override
        public int compare(pair o1, pair o2) {
            return o1.y - o2.y;
        }
    }
    public static long modularpow(long a,long b){
        long res=1;
        if(b==0)
            return res;
        else
        {
            while(b>0){
                if(b%2==1){
                    res*=a;
                    res%=mod;

                }
                a=a*a;
                a%=mod;
                b/=2;
            }
            return res%mod;
        }
    }
    public static int gcd(int a,int b){
        if(b==0)
            return a;
        else
            return gcd(b,a%b);
    }
 public static String binary(int a){
        String s1="";
        while(a>0){
            s1=a%2+s1;
            a/=2;
        }
        return s1;
    }
    public static int Lower_Bound(long a[],int l,int r,long k){


    while(r-l>1){
        int mid=l+(r-l)/2;
        if(a[mid]<=k) l=mid;
        else
            r=mid;

    }
    return l;
    }
    public static int Upper_Bound(int a[],int l,int r,int k){

        while(r-l>1){
            int mid=(l+r)/2;
            if(a[mid]<=k)
                l=mid;
            else
                r=mid;
        }
        return l+1;
    }
    public static void main(String arg[]) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter sb = new BufferedWriter(
                new OutputStreamWriter(System.out));
           int t=Integer.parseInt(br.readLine());
           while(t-->0){
               int n=Integer.parseInt(br.readLine());
               int k=0;
               String s[]=br.readLine().split(" ");
               while(k < n && Integer.parseInt(s[k]) == 1) {
                   k++;
               }
               if((k == n) ^ ((k % 2)==1) )
                   sb.write("Second\n");
               else
                   sb.write("First\n");

           }

            sb.flush();
    }
}
