description: Polycarp had an array a of 3 positive integers. He wrote out the sums of all non-empty subsequences of this array, sorted them in non-decreasing order, and got an array b of 7 integers.
  
  For example, if a = \{1, 4, 3\}, then Polycarp wrote out 1, 4, 3, 1 + 4 = 5, 1 + 3 = 4, 4 + 3 = 7, 1 + 4 + 3 = 8. After sorting, he got an array b = \{1, 3, 4, 4, 5, 7, 8\}.
  
  Unfortunately, Polycarp lost the array a. He only has the array b left. Help him to restore the array a.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
  
  Each test case consists of one line which contains 7 integers b_1, b_2, ..., b_7 (1 ≤ b_i ≤ 10^9; b_i ≤ b_{i+1}). 
  
  Additional constraint on the input: there exists at least one array a which yields this array b as described in the statement.
  
  Output
  
  For each test case, print 3 integers — a_1, a_2 and a_3. If there can be several answers, print any of them.
  
  Example
  
  Input
  
  
  5
  1 3 4 4 5 7 8
  1 2 3 4 5 6 7
  300000000 300000000 300000000 600000000 600000000 600000000 900000000
  1 1 2 999999998 999999999 999999999 1000000000
  1 2 2 3 3 4 5
  
  
  Output
  
  
  1 4 3
  4 1 2
  300000000 300000000 300000000
  999999998 1 1
  1 2 2
  
  Note
  
  The subsequence of the array a is a sequence that can be obtained from a by removing zero or more of its elements.
  
  Two subsequences are considered different if index sets of elements included in them are different. That is, the values of the elements don't matter in the comparison of subsequences. In particular, any array of length 3 has exactly 7 different non-empty subsequences.
solution: #include <bits/stdc++.h>
  using namespace std;
  using namespace chrono;
  void _print(long long t) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  long long gcd(long long a, long long b) {
    if (b > a) {
      return gcd(b, a);
    }
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  long long expo(long long a, long long b, long long mod) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b = b >> 1;
    }
    return res;
  }
  void extendgcd(long long a, long long b, long long *v) {
    if (b == 0) {
      v[0] = 1;
      v[1] = 0;
      v[2] = a;
      return;
    }
    extendgcd(b, a % b, v);
    long long x = v[1];
    v[1] = v[0] - v[1] * (a / b);
    v[0] = x;
    return;
  }
  long long mminv(long long a, long long b) {
    long long arr[3];
    extendgcd(a, b, arr);
    return arr[0];
  }
  long long mminvprime(long long a, long long b) { return expo(a, b - 2, b); }
  bool revsort(long long a, long long b) { return a > b; }
  void swap(int &x, int &y) {
    int temp = x;
    x = y;
    y = temp;
  }
  long long combination(long long n, long long r, long long m, long long *fact,
                        long long *ifact) {
    long long val1 = fact[n];
    long long val2 = ifact[n - r];
    long long val3 = ifact[r];
    return (((val1 * val2) % m) * val3) % m;
  }
  void google(int t) { cout << "Case #" << t << ": "; }
  vector<long long> sieve(int n) {
    int *arr = new int[n + 1]();
    vector<long long> vect;
    for (int i = 2; i <= n; i++)
      if (arr[i] == 0) {
        vect.push_back(i);
        for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
      }
    return vect;
  }
  long long mod_add(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long mod_mul(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long mod_sub(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  long long mod_div(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (mod_mul(a, mminvprime(b, m), m) + m) % m;
  }
  long long phin(long long n) {
    long long number = n;
    if (n % 2 == 0) {
      number /= 2;
      while (n % 2 == 0) n /= 2;
    }
    for (long long i = 3; i <= sqrt(n); i += 2) {
      if (n % i == 0) {
        while (n % i == 0) n /= i;
        number = (number / i * (i - 1));
      }
    }
    if (n > 1) number = (number / n * (n - 1));
    return number;
  }
  void solve() {
    int t;
    cin >> t;
    while (t--) {
      int b[7];
      for (int i = 0; i < 7; i++) {
        cin >> b[i];
      }
      cout << b[6] - b[5] << " " << b[1] << " " << b[6] - (b[0] + b[1]);
      cout << endl;
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    auto start1 = high_resolution_clock::now();
    solve();
    auto stop1 = high_resolution_clock::now();
    auto duration = duration_cast<microseconds>(stop1 - start1);
  }
  