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.

import java.io.*;
import java.util.*;
import java.lang.*;


public class Main {
    static ArrayList<Long> arr;

    public static void main(String[] args) throws IOException {
        FastReader sc = new FastReader();
        int t = sc.nextInt();
        while (t-- != 0) {
            int n = sc.nextInt();
            int a[] = new int[n];

            long sum = 0;
            long xor = 0;
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
                sum += a[i];
                xor ^= a[i];
            }
                System.out.println("2");
                System.out.println(xor+" "+(sum+xor));
        




        }
    }


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

    // method to return LCM of two numbers
    static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    public 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 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 long simpleSieve(int limit, Vector<Integer> prime) {
        long count = 0;
        boolean mark[] = new boolean[limit + 1];

        for (int i = 0; i < mark.length; i++)
            mark[i] = true;

        for (int p = 2; p * p < limit; p++) {
            // If p is not changed, then it is a prime
            if (mark[p] == true) {
                // Update all multiples of p
                for (int i = p * p; i < limit; i += p)
                    mark[i] = false;
            }
        }

        // Print all prime numbers and store them in prime
        for (int p = 2; p < limit; p++) {
            if (mark[p] == true) {
                prime.add(p);
                count++;
                /*System.out.print(p + "  ");*/
            }
        }
        return count;
    }

    // Prints all prime numbers smaller than 'n'
    static long segmentedSieve(long n) {
        // Compute all primes smaller than or equal
        // to square root of n using simple sieve
        int limit = (int) (Math.floor(Math.sqrt(n)) + 1);
        Vector<Integer> prime = new Vector<>();
        long count = simpleSieve(limit, prime);

        // Divide the range [0..n-1] in different segments
        // We have chosen segment size as sqrt(n).
        int low = limit;
        long high = 2 * limit;

        // While all segments of range [0..n-1] are not processed,
        // process one segment at a time
        while (low < n) {
            if (high >= n)
                high = n;

            // To mark primes in current range. A value in mark[i]
            // will finally be false if 'i-low' is Not a prime,
            // else true.
            boolean mark[] = new boolean[limit + 1];

            for (int i = 0; i < mark.length; i++)
                mark[i] = true;

            // Use the found primes by simpleSieve() to find
            // primes in current range
            for (int i = 0; i < prime.size(); i++) {
                // Find the minimum number in [low..high] that is
                // a multiple of prime.get(i) (divisible by prime.get(i))
                // For example, if low is 31 and prime.get(i) is 3,
                // we start with 33.
                int loLim = (int) (Math.floor(low / prime.get(i)) * prime.get(i));
                if (loLim < low)
                    loLim += prime.get(i);

                /*  Mark multiples of prime.get(i) in [low..high]:
                    We are marking j - low for j, i.e. each number
                    in range [low, high] is mapped to [0, high-low]
                    so if range is [50, 100]  marking 50 corresponds
                    to marking 0, marking 51 corresponds to 1 and
                    so on. In this way we need to allocate space only
                    for range  */
                for (int j = loLim; j < high; j += prime.get(i))
                    mark[j - low] = false;
            }

            // Numbers which are not marked as false are prime
            for (int i = low; i < high; i++)
                if (mark[i - low] == true) {
                    count++;
                    /*System.out.print(i + "  ");*/
                }


            // Update low and high for next segment
            low = low + limit;
            high = high + limit;
        }
        return count;
    }


}