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

public class Main {
	static final int MOD_PRIME = 1000000007;

	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		Task solver = new Task();

		int i = 0;
		int t = 1;
		t = in.nextInt();
		for (; i < t; i++)
			solver.solve(i, in, out);

		out.close();
	}

	static class Task {
		public void solve(int testNumber, InputReader in, PrintWriter out) {
			int n = in.nextInt();
			long[] ar = new long[n];
			long sum = 0;
			long l = 0;
			for(int i = 0; i < n; i++) {
				ar[i] = in.nextLong();
				l ^= ar[i];
				sum += ar[i];
			}
			
			long c = l + sum;
			out.println(2);
			out.println(l + " " + c);
			
					
		}
	}

	// template code

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

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

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

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

	}

	static long modexp(long a, long b, long p) {
		// returns a to the power b mod p by modular exponentiation

		long res = 1;
		long mult = a % p;
		while (b > 0) {
			if (b % 2 == 1) {
				res = (res * mult) % p;
			}
			b /= 2;
			mult = (mult * mult) % p;
		}

		return res;
	}

	static double log(double arg, double base) {
		// returns log of a base b, contains floating point errors, dont use for exact
		// calculations.

		if (base < 0 || base == 1) {
			throw new ArithmeticException("base cant be 1 or negative");
		}
		if (arg < 0) {
			throw new ArithmeticException("log of negative number undefined");
		}
		return Math.log10(arg) / Math.log10(base);
	}

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

		return gcd(b, a % b);

	}

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

		return gcd(b, a % b);

	}

	// scope for improvement
	static ArrayList<Integer> sieveOfEratosthenes(int n) {
		boolean[] check = new boolean[n + 1];
		ArrayList<Integer> prime = new ArrayList<Integer>();
		for (long i = 2; i <= n; i++) {
			if (!check[(int) i]) {
				prime.add((int) i);
				for (long j = i * i; j <= n; j += i) {
					check[(int) j] = true;
				}
			}
		}

		return prime;
	}

	static int modInverse(int a, int n) {
		// returns inverse of a mod n by extended euclidean algorithm
		int t = 0;
		int newt = 1;
		int r = n;
		int newr = a;
		int quotient;
		int tempr, tempt;
		while (newr != 0) {
			quotient = r / newr;
			tempt = newt;
			tempr = newr;
			newr = r - quotient * tempr;
			newt = t - quotient * tempt;
			t = tempt;
			r = tempr;
		}
		if (r > 1) {
			throw new ArithmeticException("inverse of " + a + " mod " + n + " does not exist");
		} else {
			if (t < 0) {
				t += n;
			}
			return t;
		}
	}

	static int primeModInverse(int a, int n) {
		// returns inverse of a mod n by mod exponentiation, use only if n is prime
		return (int) modexp(a, n - 2, n);
	}

	static void dfs(HashMap<Integer, ArrayList<Integer>> adj, Set<Integer> ans, Set<Integer> open,
			HashMap<String, Integer> edge, boolean[] vis, int i) {
		vis[i] = true;
		open.add(i);
		if (adj.get(i) != null) {
			for (int k : adj.get(i)) {
				if (!vis[k]) {
					dfs(adj, ans, open, edge, vis, k);
				} else if (open.contains(k)) {
					ans.add(edge.get(String.valueOf(i) + " " + String.valueOf(k)));
				}
			}
		}
		open.remove(i);
	}
}