Problem: Let's call an array a_1, a_2, ..., a_m of nonnegative integer numbers good if a_1 + a_2 + ... + a_m = 2⋅(a_1 ⊕ a_2 ⊕ ... ⊕ a_m), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

For example, array [1, 2, 3, 6] is good, as 1 + 2 + 3 + 6 = 12 = 2⋅ 6 = 2⋅ (1⊕ 2 ⊕ 3 ⊕ 6). At the same time, array [1, 2, 1, 3] isn't good, as 1 + 2 + 1 + 3 = 7 ≠ 2⋅ 1 = 2⋅(1⊕ 2 ⊕ 1 ⊕ 3).

You are given an array of length n: a_1, a_2, ..., a_n. Append at most 3 elements to it to make it good. Appended elements don't have to be different. It can be shown that the solution always exists under the given constraints. If there are different solutions, you are allowed to output any of them. Note that you don't have to minimize the number of added elements!. So, if an array is good already you are allowed to not append elements.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10 000). The description of the test cases follows.

The first line of each test case contains a single integer n (1≤ n ≤ 10^5) — the size of the array.

The second line of each test case contains n integers a_1, a_2, ..., a_n (0≤ a_i ≤ 10^9) — the elements of the array.

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case, output two lines.

In the first line, output a single integer s (0≤ s≤ 3) — the number of elements you want to append.

In the second line, output s integers b_1, ..., b_s (0≤ b_i ≤ 10^{18}) — the elements you want to append to the array.

If there are different solutions, you are allowed to output any of them.

Example

Input


3
4
1 2 3 6
1
8
2
1 1


Output


0

2
4 4
3
2 6 2

Note

In the first test case of the example, the sum of all numbers is 12, and their ⊕ is 6, so the condition is already satisfied.

In the second test case of the example, after adding 4, 4, the array becomes [8, 4, 4]. The sum of numbers in it is 16, ⊕ of numbers in it is 8.

Solution (in JAVA):

Solution:

import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.*;
import java.io.BufferedReader;
import java.io.InputStreamReader;
//import java.math.*;
 
/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
 
public class Main {
	
	static int MD= 1000000007 ;
    public static void main(String[] args) throws IOException {
    	OutputStream outputStream = System.out;
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }
   
    static class TaskC {
   
	    public void solve(int testNumber, InputReader in, PrintWriter out) throws IOException {
	    	int t=in.nextInt();
	    	while(t-->0) { 
	    		int n=in.nextInt();
	    		int[] a=new int[n];
	    		long sum=0;
	    		long xor=0;
	    		for(int i=0;i<n;i++) {
	    			a[i]=in.nextInt();
	    			sum+=a[i];
	    			xor^=a[i];
	    		}
	    		if(sum!=2*xor) {
	    			out.println(2);
		    		out.println(xor+" "+(sum+xor));
	    		}
	    		else {
	    			out.println(0);
	    			out.println();
	    		}
	    		
	    		
	    		
	    		
	    		
	        }
	 
	    	
	    }
	   
    
    }
   
   
 
 
    
    static class InputReader
    {
        BufferedReader br;
        StringTokenizer st;
 
        public InputReader()
        {
            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 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;
        }
    
    }
    static void sort(int[] a, int low, int high) 
    {
        int N = high - low;         
        if (N <= 1) 
            return; 
        int mid = low + N/2; 
        sort(a, low, mid); 
        sort(a, mid, high); 
        int[] temp = new int[N];
        int i = low, j = mid;
        for (int k = 0; k < N; k++) 
        {
            if (i == mid)  
                temp[k] = a[j++];
            else if (j == high) 
                temp[k] = a[i++];
            else if (a[j]<a[i]) 
                temp[k] = a[j++];
            else 
                temp[k] = a[i++];
        }    
        for (int k = 0; k < N; k++) 
            a[low + k] = temp[k];         
    }
    static void lsort(long[] a, int low, int high) 
    {
        int N = high - low;         
        if (N <= 1) 
            return; 
        int mid = low + N/2; 
        lsort(a, low, mid); 
        lsort(a, mid, high); 
        long[] temp = new long[N];
        int i = low, j = mid;
        for (int k = 0; k < N; k++) 
        {
            if (i == mid)  
                temp[k] = a[j++];
            else if (j == high) 
                temp[k] = a[i++];
            else if (a[j]<a[i]) 
                temp[k] = a[j++];
            else 
                temp[k] = a[i++];
        }    
        for (int k = 0; k < N; k++) 
            a[low + k] = temp[k];         
    }
    public static long fexp(long x, long y, long p) {
        long res = 1;      
        
       
        x = x % p;  
      
        while (y > 0) 
        { 
         
            if((y & 1)==1) 
                res = (res * x) % p; 
      
          
            y = y >> 1;  
            x = (x * x) % p;  
        } 
        return res; 
        
    }
    static int gcd(int a, int b) 
    { 
        if (b == 0) 
            return a; 
        return gcd(b, a % b);  
          
    }
    static long lgcd(long a, long b) 
    { 
        if (b == 0) 
            return a; 
        return lgcd(b, a % b);  
          
    }
  
}