Problem: Chaneka has a hobby of playing with animal toys. Every toy has a different fun value, a real number. Chaneka has four boxes to store the toys with specification: 

  * The first box stores toys with fun values in range of (-∞,-1]. 
  * The second box stores toys with fun values in range of (-1, 0). 
  * The third box stores toys with fun values in range of (0, 1). 
  * The fourth box stores toys with fun value in range of [1, ∞). 



Chaneka has A, B, C, D toys in the first, second, third, and fourth box, respectively. One day she decides that she only wants one toy, a super toy. So she begins to create this super toy by sewing all the toys she has.

While the number of toys Chaneka has is more than 1, she takes two different toys randomly and then sews them together, creating a new toy. The fun value of this new toy is equal to the multiplication of fun values of the sewn toys. She then puts this new toy in the appropriate box. She repeats this process until she only has one toy. This last toy is the super toy, and the box that stores this toy is the special box.

As an observer, you only know the number of toys in each box initially but do not know their fun values. You also don't see the sequence of Chaneka's sewing. Determine which boxes can be the special box after Chaneka found her super toy.

Input

The first line has an integer T (1 ≤ T ≤ 5 ⋅ 10^4), the number of test cases.

Every case contains a line with four space-separated integers A B C D (0 ≤ A, B, C, D ≤ 10^6, A + B + C + D > 0), which denotes the number of toys in the first, second, third, and fourth box, respectively.

Output

For each case, print four space-separated strings. Each string represents the possibility that the first, second, third, and fourth box can be the special box from left to right.

For each box, print "Ya" (Without quotes, Indonesian for yes) if that box can be the special box. Print "Tidak" (Without quotes, Indonesian for No) otherwise.

Example

Input


2
1 2 0 1
0 1 0 0


Output


Ya Ya Tidak Tidak
Tidak Ya Tidak Tidak

Note

For the first case, here is a scenario where the first box is the special box: 

  * The first box had toys with fun values \{-3\}. 
  * The second box had toys with fun values \{ -0.5, -0.5 \} 
  * The fourth box had toys with fun values \{ 3 \} 



The sewing sequence: 

  1. Chaneka sews the toy with fun -0.5 and -0.5 to a toy with fun 0.25 and then put it in the third box. 
  2. Chaneka sews the toy with fun -3 and 0.25 to a toy with fun -0.75 and then put it in the second box. 
  3. Chaneka sews the toy with fun -0.75 and 3 to a toy with fun -1.25 and then put it in the first box, which then became the special box. 



Here is a scenario where the second box ends up being the special box: 

  * The first box had toys with fun values \{-3\} 
  * The second box had toys with fun values \{ -0.33, -0.25 \}. 
  * The fourth box had toys with fun values \{ 3 \}. 



The sewing sequence: 

  1. Chaneka sews the toy with fun -3 and -0.33 to a toy with fun 0.99 and then put it in the third box. 
  2. Chaneka sews the toy with fun 0.99 and 3 to a toy with fun 2.97 and then put in it the fourth box. 
  3. Chaneka sews the toy with fun 2.97 and -0.25 to a toy with fun -0.7425 and then put it in the second box, which then became the special box. 

There is only one toy for the second case, so Chaneka does not have to sew anything because that toy, by definition, is the super toy.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long LLINF = 1e18;
const int INF = INT_MAX;
const long double PI = acos(-1);
const long double E = 2.71828183;
const int MOD = 1e9 + 7;
template <typename T>
struct segment_tree {
  vector<T> tree;
  int size;
  T DEF;
  void init(int n) {
    size = 1;
    DEF = INF;
    while (size < n) size *= 2;
    tree = vector<T>(2 * size, 0);
  }
  T merge(T a, T b) { return a + b; }
  void set(int i, int x, T v, int lx, int rx) {
    if (rx - lx == 1) {
      tree[x] = v;
      return;
    }
    int m = (rx + lx) / 2;
    if (i < m) {
      set(i, 2 * x + 1, v, lx, m);
    } else {
      set(i, 2 * x + 2, v, m, rx);
    }
    tree[x] = merge(tree[2 * x + 1], tree[2 * x + 2]);
  }
  void set(int i, T v) { set(i, 0, v, 0, size); }
  T sum(int l, int r, int x, int lx, int rx) {
    if (r <= lx || rx <= l)
      return 0;
    else if (l <= lx && rx <= r)
      return tree[x];
    int m = (lx + rx) / 2;
    return merge(sum(l, r, 2 * x + 1, lx, m), sum(l, r, 2 * x + 2, m, rx));
  }
  T sum(int l, int r) { return sum(l, r + 1, 0, 0, size); }
  void print_tree() {
    for (int i = 0; i < tree.size(); i++) cout << to_string(tree[i]) << " ";
    cout << "\n";
  }
};
vector<int> bfs(vector<vector<int> > adj, int src) {
  queue<int> q;
  vector<int> d(adj.size(), INF);
  q.push(src);
  d[src] = 0;
  while (!q.empty()) {
    int c = q.front();
    q.pop();
    for (int nbr : adj[c]) {
      if (d[nbr] == INF) q.push(nbr);
      d[nbr] = min(d[nbr], d[c] + 1);
    }
  }
  return d;
}
template <typename T>
void read(vector<T>& v) {
  int n = v.size();
  for (int i = 0; i < n; i++) cin >> v[i];
}
long double log(long double a, long double b) { return log(b) / log(a); }
long long power(long long base, long long exp, long long M = LLONG_MAX) {
  long long res = 1;
  while (exp) {
    if (exp % 2 == 1) res = ((res % M) * (base % M)) % M;
    base = ((base % M) * (base % M)) % M;
    exp /= 2;
  }
  return res;
}
string to_base(int n, int new_base) {
  string s;
  int nn = n;
  while (nn) {
    s += to_string(nn % new_base);
    nn /= new_base;
  }
  reverse(s.begin(), s.end());
  return s;
}
long long gcd(long long a, long long b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long x = (a / gcd(a, b));
  return b * x;
}
bitset<(const int)(5e5)> sieve() {
  bitset<(const int)(5e5)> ans;
  for (int i = 2; i < 5e5; i++) ans[i] = 1;
  for (int i = 2; i * i < 5e5; i++) {
    if (ans[i])
      for (int j = 2 * i; j < 5e5; j += i) ans[j] = 0;
  }
  return ans;
}
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
void solve() {
  int a, b, c, d;
  cin >> a >> b >> c >> d;
  if ((a + b) % 2 == 0) {
    if ((b || c) && !a && !d) {
      cout << "Tidak Tidak Ya Tidak\n";
      return;
    }
    if ((a || d) && (b || c))
      cout << "Tidak Tidak Ya Ya\n";
    else
      cout << "Tidak Tidak Tidak Ya\n";
  } else {
    if ((b || c) && !a && !d) {
      cout << "Tidak Ya Tidak Tidak\n";
      return;
    }
    if ((b || c) && (a || d))
      cout << "Ya Ya Tidak Tidak\n";
    else
      cout << "Ya Tidak Tidak Tidak\n";
  }
}