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 JAVA):

Solution:


 //   * * * its fun to do the impossible * * *   //
 import java.io.BufferedReader;
 import java.io.IOException;
 import java.io.InputStreamReader;
 import java.util.*;
  
 
 public class HugeBozesOfAnimalToys {
     static class Pair implements Comparable<Pair>{
         int a;
         int b;
 
         Pair(int a , int b){
             this.a = a;
             this.b = b;
         }
 
         public int compareTo(Pair o){
             return this.a - o.a;
         }
     }
    
     public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int t = sc.nextInt();
        
        while(t-->0){
            long a = sc.nextLong();
            long b = sc.nextLong();
            long c = sc.nextLong();
            long d = sc.nextLong();

            String res[] = new String[4];
            for(int i=0;i<4;i++){
                res[i] = "Tidak";
            }

            if(((a + b) & 1) == 1){
                if(b > 0 || c > 0){
                    res[1] = "Ya";
                }

                if(a > 0 || d > 0){
                    res[0] = "Ya";
                }
            }
            else{
                if(b > 0 || c > 0){
                    res[2] = "Ya";
                }
                
                if(d > 0 || a > 0){
                    res[3] = "Ya";
                }
            }

            for(String e : res){
                System.out.print(e+ " ");
            }
            System.out.println();
        }
     }
 
 
 
 
 
 
  
     // Use this instead of Arrays.sort() on an array of ints. Arrays.sort() is n^2
     // worst case since it uses a version of quicksort. Although this would never
     // actually show up in the real world, in codeforces, people can hack, so
     // this is needed.
     static void ruffleSort(int[] a) {
         //ruffle
         int n=a.length;
         Random r=new Random();
         for (int i=0; i<a.length; i++) {
             int oi=r.nextInt(n), temp=a[i];
             a[i]=a[oi];
             a[oi]=temp;
         }
         
         //then sort
         Arrays.sort(a);
     }
     
     // Use this to input code since it is faster than a Scanner
     static class FastScanner {
         BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
         StringTokenizer st=new StringTokenizer("");
         String next() {
             while (!st.hasMoreTokens())
                 try {
                     st=new StringTokenizer(br.readLine());
                 } catch (IOException e) {
                     e.printStackTrace();
                 }
             return st.nextToken();
         }
         
         int nextInt() {
             return Integer.parseInt(next());
         }
 
         double nextDouble() {
             return Double.parseDouble(next());
         }
  
         String str = "";
  
         String nextLine() {
             try {
                 str = br.readLine();
             } catch (IOException e) {
                 e.printStackTrace();
             }
             return str;
         }
  
         int[] readArray(int n) {
             int[] a = new int[n];
             for (int i = 0; i < n; i++)
                 a[i] = nextInt();
             return a;
         }
  
         long[] readLongArray(int n) {
             long[] a = new long[n];
             for (int i = 0; i < n; i++)
                 a[i] = nextLong();
             return a;
         }
         long nextLong() {
             return Long.parseLong(next());
         }
     }
     // use this to find the index of any element in the array +1 /// 
     // returns an array that corresponds to the index of the i+1th in the array a[]
     // runs only for array containing different values enclosed btw 0 to n-1
     static int[] indexOf(int[] a) {
         int[] toRet=new int[a.length];
         for (int i=0; i<a.length; i++) {
             toRet[a[i]]=i+1;
         }
         return toRet;
     }
 
     static int gcd(int a, int b) {
         if (b==0) return a;
         return gcd(b, a%b);
     }
 
     //generates all the prime numbers upto n
     static void sieveOfEratosthenes(int n , ArrayList<Integer> al)
     {
         boolean prime[] = new boolean[n + 1];
         for (int i = 0; i <= n; i++)
             prime[i] = true;
  
         for (int p = 2; p * p <= n; p++) 
         {
             if (prime[p] == true) 
             {
                 for (int i = p * p; i <= n; i += p)
                     prime[i] = false;
             }
         }
         for (int i = 2; i <= n; i++)
         {
             if (prime[i] == true)
                 al.get(i);
         }
     }
     static final int mod=100000000 + 7;
     
     //fastPow
     static long fastPow(long base, long exp) {
         if (exp==0) return 1;
         long half=fastPow(base, exp/2);
         if (exp%2==0) return mul(half, half);
         return mul(half, mul(half, base));
     }
 
     //multiply two long numbers
     static long mul(long a, long b) {
         return a*b%mod;
     }
 
     static int nCr(int n, int r)
     {
            return fact(n) / (fact(r) *
                 fact(n - r));
     }
     
     // Returns factorial of n
     static int fact(int n)
     {
         int res = 1;
         for (int i = 2; i <= n; i++)
             res = res * i;
         return res;
     }
      
     // to generate the lps array
     // lps means longest preffix that is also a suffix
     static void generateLPS(int lps[] , String p){
         int l = 0;
         int r = 1;
    
         while(l < p.length() && l < r && r < p.length()){
             if(p.charAt(l) == p.charAt(r)){
                 lps[r] = l + 1;
                 l++;
                 r++;
             }
             else{
                 if(l > 0)
                     l = lps[l - 1];
                 else
                     r++;    
             }
         }
     }
      
 
     //count sort --> it runs in O(n) time but compromises in space
     static ArrayList<Integer> countSort(int a[]){
         int max = Integer.MIN_VALUE;
         for(int i=0;i<a.length;i++){
             max = Math.max(max , a[i]);
         }
 
         int posfre[] = new int[max+1];
         boolean negPres = false;
         for(int i=0;i<a.length;i++){
             if(a[i]>=0){
                 posfre[a[i]]++;
             }
             else{
                 negPres = true;
             }
         }
         ArrayList<Integer> res = new ArrayList<>();
         if(negPres){
             int min = Integer.MAX_VALUE;
             for(int i=0;i<a.length;i++){
                 min = Math.min(min , a[i]);
             }
 
             int negfre[] = new int[-1*min+1];
             for(int i=0;i<a.length;i++){
                 if(a[i]<0){
                     negfre[-1*a[i]]++;
                 }
             }
 
             for(int i=min;i<0;i++){
                 for(int j=0;j<negfre[-1*i];j++){
                     res.add(i);
                 }
             }
             for(int i=0;i<=max;i++){
                 for(int j=0;j<posfre[i];j++){
                     res.add(i);
                 }
             }
             return res;
         }
         for(int i=0;i<=max;i++){
             for(int j=0;j<posfre[i];j++){
                 res.add(i);
             }
         }
         return res;
     }
 
     // a -> z == 97 -> 122
 
     // String.format("%.9f", ans) ,--> to get upto 9 decimal places , (ans is double)
     
     // write 
 }