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;
struct BigNum {
  vector<long long> value;
  void set(long long x) {
    value = *(new vector<long long>);
    value.push_back(x);
  }
  void duplicate(BigNum other) { value = other.value; }
  void add(BigNum other) {
    vector<long long> o = other.value;
    if (o.size() > value.size()) {
      vector<long long> t = value;
      value = o;
      o = t;
    }
    for (int i = 0; i < o.size(); i += 1) {
      value[i] += o[i];
      if (i < value.size() - 1 && value[i] >= (long long)1e9) {
        value[i + 1] += value[i] / (long long)1e9;
        value[i] = value[i] % (long long)1e9;
      }
    }
    if (value.back() >= (long long)1e9) {
      value.push_back(value.back() / (long long)1e9);
      value[value.size() - 2] = value[value.size() - 2] % (long long)1e9;
    }
  }
  void mul(long long mul) {
    for (int i = 0; i < value.size(); i += 1) {
      value[i] *= mul;
    }
    for (int i = 0; i < value.size() - 1; i += 1) {
      if (value[i] >= (long long)1e9) {
        value[i + 1] += value[i] / (long long)1e9;
        value[i] = value[i] % (long long)1e9;
      }
    }
    if (value.back() >= (long long)1e9) {
      value.push_back(value.back() / (long long)1e9);
      value[value.size() - 2] = value[value.size() - 2] % (long long)1e9;
    }
  }
  long double div(BigNum other1, BigNum other2) {
    int s = value.size();
    long double current = value.back();
    if (s >= 2) {
      current += ((long double)value[s - 2]) / (long long)1e9;
    }
    if (s >= 3) {
      current +=
          ((long double)value[s - 3]) / ((long long)1e9 * (long long)1e9);
    }
    int s1 = other1.value.size();
    int s2 = other2.value.size();
    long double other = 0;
    if (s1 + s2 - 1 == s) {
      other += other1.value.back() * other2.value.back();
      if (s1 >= 2) {
        other += ((long double)other1.value[s1 - 2] * other2.value.back()) /
                 (long long)1e9;
      }
      if (s2 >= 2) {
        other += ((long double)other2.value[s2 - 2] * other1.value.back()) /
                 (long long)1e9;
      }
      if (s1 >= 3) {
        other += ((long double)other1.value[s1 - 3] * other2.value.back()) /
                 ((long long)1e9 * (long long)1e9);
      }
      if (s2 >= 3) {
        other += ((long double)other2.value[s2 - 3] * other1.value.back()) /
                 ((long long)1e9 * (long long)1e9);
      }
    }
    if (s1 + s2 - 1 == s - 1) {
      other += other1.value.back() * other2.value.back();
      if (s1 >= 2) {
        other += ((long double)other1.value[s1 - 2] * other2.value.back()) /
                 (long long)1e9;
      }
      if (s2 >= 2) {
        other += ((long double)other2.value[s2 - 2] * other1.value.back()) /
                 (long long)1e9;
      }
      other /= (long long)1e9;
    }
    if (s1 + s2 - 1 == s - 2) {
      other += other1.value.back() * other2.value.back();
      other /= (long long)1e9 * (long long)1e9;
    }
    return other / current;
  }
  string stringify() {
    string out = to_string(value.back());
    for (int i = value.size() - 1 - 1; i > -1; i -= 1) {
      string cur = to_string(value[i] + (long long)1e9);
      out += cur.substr(1);
    }
    return out;
  }
};
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long triangle(long long x) { return (((long long)(x + 1)) * (x)) / 2; }
long long modInverse(long long a) {
  long long m = 1000000007;
  long long y = 0, x = 1;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) {
    x += 1000000007;
  }
  return x;
}
long long modInverse(long long a, long long b) {
  long long m = b;
  long long y = 0, x = 1;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) {
    x += b;
  }
  return x;
}
long long pow(long long a, long long b) {
  if (a <= 1) {
    return a;
  }
  if (b == 0) {
    return 1;
  }
  if (b % 2 == 0) {
    return pow((a * a) % 1000000007, b / 2) % 1000000007;
  }
  return (a * pow((a * a) % 1000000007, b / 2)) % 1000000007;
}
vector<long long> dev(vector<pair<long long, long long> > divisors) {
  if (divisors.size() == 0) {
    vector<long long> cur;
    cur.push_back(1);
    return cur;
  }
  long long x = divisors.back().first;
  long long n = divisors.back().second;
  divisors.pop_back();
  vector<long long> ans = dev(divisors);
  vector<long long> cur;
  long long xi = 1;
  for (int i = 0; i < n + 1; i += 1) {
    for (int j = 0; j < ans.size(); j += 1) {
      cur.push_back(ans[j] * xi);
    }
    xi *= x;
  }
  return cur;
}
void add(vector<int>& x, vector<int>& y) {
  for (int i = 0; i < x.size(); i += 1) {
    x[i] += y[i];
  }
}
struct pt {
  long long x, y;
  long long d() const { return x * x + y * y; }
};
inline pt operator-(const pt& a) { return {-a.x, -a.y}; }
inline pt operator+(const pt& a, const pt& b) { return {a.x + b.x, a.y + b.y}; }
inline pt operator-(const pt& a, const pt& b) { return {a.x - b.x, a.y - b.y}; }
inline long long operator*(const pt& a, const pt& b) {
  return a.x * b.y - a.y * b.x;
}
inline bool operator<(const pt& a, const pt& b) { return a * b < 0; }
bool llll(const pt& a, const pt& b) { return a * b > 0; }
inline bool operator<=(const pt& a, const pt& b) { return a * b <= 0; }
inline long long operator/(const pt& a, const pt& b) {
  return a.x * b.x + a.y * b.y;
}
inline bool operator==(const pt& a, const pt& b) {
  return a.x == b.x && a.y == b.y;
}
inline bool operator!=(const pt& a, const pt& b) {
  return a.x != b.x || a.y != b.y;
}
pt o = {0, 0};
bool comp(pt a, pt b) { return (a - o) * (b - o) < 0; }
bool isprime(long long a) {
  if (a == 1) {
    return false;
  }
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  for (int kk = 0; kk < t; kk += 1) {
    long long n;
    cin >> n;
    long long sum = 0, xoor = 0;
    for (int i = 0; i < n; i += 1) {
      long long a;
      cin >> a;
      sum += a;
      xoor ^= a;
    }
    if (sum == 2 * xoor) {
      cout << 0 << endl;
      continue;
    }
    if (sum > 2 * xoor || (sum & 1)) {
      cout << 3 << endl;
      long long b = (((long long)1) << 54) + (sum & 1);
      cout << (b) << ' ';
      sum += b;
      xoor ^= b;
    } else {
      cout << 2 << endl;
    }
    cout << ((xoor * 2 - sum) / 2) << ' ';
    cout << ((xoor * 2 - sum) / 2) << endl;
  }
  return 0;
}