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.*;
import java.math.*;

public class Solution {

    static BufferedReader in;
    static StringTokenizer stk;
    static int T, N, M, K;
    static PrintWriter out;

    public static void main(String[] args) throws Exception {
        out = new PrintWriter(new OutputStreamWriter(System.out));

        in = new BufferedReader(new InputStreamReader(System.in));
        // Start of User Code

        T = ni();

        for (int i = 0; i < T; i++) {
            out.println(proc());
        }

        // End of User Code
        out.flush();
        in.close();
    }

    static String proc() throws Exception {
        int n = ni();
        long sum = 0;
        long xor = 0;
        for(int i = 0; i< n; i++){
            int t = ni();
            xor ^= t;
            sum += t;
        }
        if(sum == 2 * xor) {
            return "0\n";
        }
        if(xor == 0){
            return "1\n" + sum;
        }
        return "2\n" + (xor) + " " + (xor + sum);
    }

    static long modPow(long n, long pow, long mod) {
        return BigInteger.valueOf(n).modPow(BigInteger.valueOf(pow), BigInteger.valueOf(mod)).longValue();
    }

    static long modInv(long n, long mod, boolean isPrimeModuli) {
        if (isPrimeModuli) {
            return modPow(n, mod - 2, mod);
        }
        return BigInteger.valueOf(n).modInverse(BigInteger.valueOf(mod)).longValue();
    }
    // calc factorials
    static long[] fact;

    static void calcFactorials(int n) {
        fact = new long[n + 1];
        fact[0] = 1;
        for (int i = 1; i <= n; i++) {
            fact[i] = fact[i - 1] * i;
        }
    }

    static void calcFactorialsModM(int n, long M) {
        fact = new long[n + 1];
        fact[0] = 1;
        for (int i = 1; i <= n; i++) {
            fact[i] = fact[i - 1] * i;
            fact[i] %= M;
        }
    }

    static long ncr(int n, int r) {
        return fact[n] / (fact[n - r] * fact[r]);
    }

    static long ncrModM(int n, int r, long MOD, boolean isPrimeModuli) {
        return (((fact[n] * modInv(fact[n - r], MOD, isPrimeModuli))%MOD)* modInv(fact[r], MOD, isPrimeModuli))%MOD;
    }

    static long toL(String s) {
        return Long.parseLong(s);
    }

    static long toL(BigInteger n) {
        return n.longValue();
    }

    static int toI(String s) {
        return Integer.parseInt(s);
    }

    static double toD(String s) {
        return Double.parseDouble(s);
    }

    static void printf(String format, Object... args) {
        out.printf(format, args);
    }
    

    static int ni() throws Exception {
        while (stk == null || !stk.hasMoreTokens()) {
            stk = new StringTokenizer(in.readLine());
        }
        return Integer.parseInt(stk.nextToken());
    }

    static long nl() throws Exception {
        while (stk == null || !stk.hasMoreTokens()) {
            stk = new StringTokenizer(in.readLine());
        }
        return Long.parseLong(stk.nextToken());
    }

    static double nd() throws Exception {
        while (stk == null || !stk.hasMoreTokens()) {
            stk = new StringTokenizer(in.readLine());
        }
        return Double.parseDouble(stk.nextToken());
    }

    static String ns() throws Exception {
        while (stk == null || !stk.hasMoreTokens()) {
            stk = new StringTokenizer(in.readLine());
        }
        return stk.nextToken();
    }

    static int min(int a, int b) {
        return a < b ? a : b;
    }

    static int max(int a, int b) {
        return a > b ? a : b;
    }

    static long min(long a, long b) {
        return a < b ? a : b;
    }

    static long max(long a, long b) {
        return a > b ? a : b;
    }

    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);
    }
}