每次操作都是乘法，所以最终的结果的符号可以确定，并且和操作顺序没有关系。确定了符号之后就只有两种选择：如果乘积为正，结果可能是 $C$ 或者 $D$。接下来只需要考虑所有数字的绝对值的乘积的范围，如果绝对值中包含 $(0, 1)$ 中的数，那么结果可能是 $C$；如果绝对值中包含 $[1, \infin)$ 中的数，那么结果可能是 $D$。乘积为负同理。

// Date: Mon Dec 11 22:53:18 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())

#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

int t, a, b, c, d;
bool res[10];

int main(void) {
#ifdef _DEBUG
  freopen("1425h.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> a >> b >> c >> d;

    int sum = a + b, sum1 = b + c, sum2 = a + d;
    memset(res, false, sizeof res);

    if (sum & 1) {
      if (sum1) {
        res[2] = true;
      }
      if (sum2)
        res[1] = true;
    } else {
      if (sum1)
        res[3] = true;
      if (sum2)
        res[4] = true;
    }

    For1(i, 1, 4) { cout << (res[i] ? "Ya" : "Tidak") << ' '; }
    cout << '\n';
  }

  return 0;
}