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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
void c_p_c() {}
class DisjointSet {
  unordered_map<int, int> parent;

 public:
  void makeSet(int N) {
    for (int i = 0; i < N; i++) parent[i] = i;
  }
  int Find(int v) {
    if (v == parent[v]) return v;
    return parent[v] = Find(parent[v]);
  }
  void Union(int a, int b) {
    int x = Find(a);
    int y = Find(b);
    parent[x] = y;
  }
};
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void PrimeFactors(long long n) {
  while (n % 2 == 0) {
    cout << 2 << " ";
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    while (n % i == 0) {
      cout << i << " ";
      n = n / i;
    }
  }
  if (n > 2) cout << n << " ";
}
template <typename T>
void parray(T a[], T n) {
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
vector<long long> uniqueelement(vector<long long> &a) {
  auto ip = unique((a).begin(), (a).end());
  a.resize(distance(a.begin(), ip));
  return a;
}
bool IsPrime(long long x) {
  if (x == 1) return false;
  for (int i = 2; 1ll * i * i <= x; i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
void leftrotate(string &s, int d) {
  reverse(s.begin(), s.begin() + d);
  reverse(s.begin() + d, s.end());
  reverse(s.begin(), s.end());
}
void rightrotate(string &s, int d) { leftrotate(s, s.length() - d); }
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
int XX[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int YY[] = {-1, 0, 1, -1, 1, -1, 0, 1};
vector<long long> sieve() {
  vector<long long> primes;
  bool a[1000000] = {0};
  int mprimes = 1000000;
  a[0] = a[1] = true;
  for (int i = 2; i * i < mprimes; ++i) {
    if (a[i] == false) {
      for (int j = i * i; j < mprimes; j = j + i) {
        a[j] = true;
      }
    }
  }
  for (int i = 2; i < mprimes; ++i) {
    if (!a[i]) primes.push_back(i);
    ;
  }
  return primes;
}
void Ynot() {
  long long i, n, m, k, c = 0, sum = 0;
  bool flag = false;
  cin >> n;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  ;
  long long x = 0;
  for (int i = 0; i < n; ++i) {
    sum += a[i];
    x = x ^ a[i];
  }
  cout << 2 << "\n";
  cout << x << " " << sum + x << "\n";
}
int main() {
  c_p_c();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 0; i < t; i++) {
    Ynot();
  }
  return 0;
}