因为操作不会影响元素的顺序，所以可以把数组排序，经过一系列操作，使得最小值和最大值相等。设最小值和最大值分别是 $a, b$，考虑 $b - a$ 的值经过一次操作后的变化 $\lfloor\frac{b + x}{2}\rfloor - \lfloor \frac{a + x}{2} \rfloor$，枚举 $a, b$ 的奇偶性，观察发现 $b - a$ 最好情况下变成 $\frac{b - a}{2}$，当 $a$ 为奇数时，令 $x = 1$，否则令 $x = 0$。

// Date: Mon Dec 25 23:28:12 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

template <typename t> istream &operator>>(istream &in, vector<t> &vec) {
  for (t &x : vec)
    in >> x;
  return in;
}

template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) {
  int n = SZ(vec);
  For(i, 0, n) {
    out << vec[i];
    if (i < n - 1)
      out << ' ';
  }
  return out;
}

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 200010;
int t, n, a[N];

int main(void) {
#ifdef _DEBUG
  freopen("1901c.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n;

    int miv = INF, mav = -INF;
    For1(i, 1, n) {
      cin >> a[i];
      miv = min(miv, a[i]);
      mav = max(mav, a[i]);
    }

    VI res;
    while (miv != mav) {
      if (miv & 1) {
        res.pb(1);
        miv = (miv + 1) / 2;
        mav = (mav + 1) / 2;
      } else {
        res.pb(0);
        miv /= 2;
        mav /= 2;
      }
    }

    int len = SZ(res);
    cout << len << '\n';
    if (len <= n && len) {
      cout << res << '\n';
    }
  }

  return 0;
}