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.*;
public class a{

    public static void main(final String [] args) throws IOException{
	   BufferInput sc=new BufferInput();
	   int t=sc.nextInt();
	   while(t-->0){
		   int n=sc.nextInt();
		   int [] a=sc.nextIntArray(n);
		   boolean f=true;

		   for(int i=n-2;i>=0;i--){
			   if(a[i]>1){
				   if(!f)f=true;
			   }
			   else{
				   f=!f;
			   }
		   }

		   if(f){
			   System.out.println("First");
		   }
		   else{
			   System.out.println("Second");
		   }

	   }
	   
	}
}


























































































  
 
 
class BufferInput 
 
{	
	final private int BUFFER_SIZE = 1 << 16;
	
	private DataInputStream din;
	
	private byte[] buffer;
	
	private int bufferPointer, bytesRead;
	
	public BufferInput() {
		din = new DataInputStream(System.in);
		buffer = new byte[BUFFER_SIZE];
		bufferPointer = bytesRead = 0;
	}
 
	public BufferInput(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];
		int cnt = 0, c;
		while ((c = read()) != -1) {
			if (c == '\n')
				break;
			buf[cnt++] = (byte) c;
		}
		return new String(buf, 0, cnt);
	}
    
    public String next() throws IOException{
 
		byte c = read();
		while(Character.isWhitespace(c)){
			c = read();
		}
		
		StringBuilder builder = new StringBuilder();
		builder.append((char)c);
		c = read();
		while(!Character.isWhitespace(c)){
			builder.append((char)c);
			c = read();
		}
		
		return builder.toString();
	}
 
	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 int[] nextIntArray(int n) throws IOException {
		int arr[] = new int[n];
		for(int i = 0; i < n; i++){
			arr[i] = nextInt();
		}
		return arr;
	}
 
	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 long[] nextLongArray(int n) throws IOException {
		long arr[] = new long[n];
		for(int i = 0; i < n; i++){
			arr[i] = nextLong();
		}
		return arr;
	}
 
	public char nextChar() throws IOException{
		byte c = read();
		while(Character.isWhitespace(c)){
			c = read();
		}
		return (char) c;	
	}
	
	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;
    }
    
	public double[] nextDoubleArray(int n) throws IOException {
		double arr[] = new double[n];
		for(int i = 0; i < n; i++){
			arr[i] = nextDouble();
		}
		return arr;
	}
 
	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();
	}
 
}