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 PYTHON3):

Solution:
##############--->>>>> Deepcoder Amit Kumar Bhuyan <<<<<---##############

"""
 Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
from __future__ import division, print_function
 
import os,sys
from io import BytesIO, IOBase
 
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
 
 
def ii():  return int(input())
def si():  return input()
def mi():  return map(int,input().strip().split(" "))
def msi(): return map(str,input().strip().split(" "))
def li():  return list(mi())
 
def dmain():
    sys.setrecursionlimit(1000000)
    threading.stack_size(1024000)
    thread = threading.Thread(target=main)
    thread.start()
    
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import log,sqrt,factorial,cos,tan,sin,radians
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *
#import threading
#from itertools import permutations
#Copy 2D list  m = [x[:] for x in mark] .. Avoid Using Deepcopy
import sys
input = sys.stdin.readline
scanner = lambda: int(input())
string = lambda: input().rstrip()
get_list = lambda: list(read())
read = lambda: map(int, input().split())
get_float = lambda: map(float, input().split())
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;

 
def ceil(x):
    if x != int(x):
        x = int(x) + 1
    return x
 
def factorial(x, m):
	val = 1
	while x>0:
		val = (val * x) % m
		x -= 1
	return val

def fact(x):
	val = 1
	while x > 0:
		val *= x
		x -= 1
	return val
    
# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp;
 
## gcd function
def gcd(a,b):
    if b == 0:
        return a;
    return gcd(b, a % b);

## lcm function
def lcm(a, b):
	return (a * b) // math.gcd(a, b)

def is_integer(n):
	return math.ceil(n) == math.floor(n)
 
## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
	if k > n:
		return 0
	if(k > n - k):
		k = n - k
	res = 1
	for i in range(k):
		res = res * (n - i)
		res = res / (i + 1)
	return int(res)
 
## upper bound function code -- such that e in a[:i] e < x;

 
## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0 and n > 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0 and n > 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes
 
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
    if y == 0:
        return 1
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 
 
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b;
 
# find function with path compression included (recursive)
# def find(x, link):
#     if link[x] == x:
#         return x
#     link[x] = find(link[x], link);
#     return link[x];
 
# find function with path compression (ITERATIVE)
def find(x, link):
    p = x;
    while( p != link[p]):
        p = link[p];
    
    while( x != p):
        nex = link[x];
        link[x] = p;
        x = nex;
    return p;
 
 
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x
 
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    prime[0], prime[1] = False, False
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime

# Euler's Toitent Function phi
def phi(n) : 
  
    result = n 
    p = 2
    while(p * p<= n) : 
        if (n % p == 0) : 
            while (n % p == 0) : 
                n = n // p 
            result = result * (1.0 - (1.0 / (float) (p))) 
        p = p + 1
    if (n > 1) : 
        result = result * (1.0 - (1.0 / (float)(n))) 
   
    return (int)(result) 

def is_prime(n):
	if n == 0:
		return False
	if n == 1:
		return True
	for i in range(2, int(n ** (1 / 2)) + 1):
		if not n % i:
			return False
 
	return True

def next_prime(n, primes):
	while primes[n] != True:
		n += 1
	return n

 
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array
 
################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
    res = []
    for i in range(2, int(x ** 0.5) + 1):
    	while x % i == 0:
    		res.append(i)
    		x //= i
    if x != 1:
   		res.append(x)
    return res
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

def factors(n):
	res = []
	for i in range(1, int(n ** 0.5) + 1):
		if n % i == 0:
			res.append(i)
			res.append(n // i)
	return list(set(res))
 
## taking integer array input
def int_array():
    return list(map(int, input().strip().split()));
 
def float_array():
    return list(map(float, input().strip().split()));
 
## taking string array input
def str_array():
    return input().strip().split();

def binary_search(low, high, w, h, n):
	while low < high:
		mid = low + (high - low) // 2
		# print(low, mid, high)
		if check(mid, w, h, n):
			low = mid + 1
		else:
			high = mid
	return low

## for checking any conditions
def check(moves, n):
	val = (moves + 1) // 2
	rem = moves - val
	sol = (val + 1) * (rem + 1)
	return sol < n
	

## for sorting according to second position
def sortSecond(val):
	return val[1]
	

 
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
alphs = "abcdefghijklmnopqrstuvwxyz"
 
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
 
from itertools import permutations
import math
import bisect as bis
import random
import sys
import collections as collect
import functools as fnt
from decimal import Decimal
# from sys import stdout
# import numpy as np
"""
_______________
rough work here
_______________

001110001

"""



def solve():
    a, b, c, d = read()
    if (a + b) % 2 == 0:
        if b + c > 0 and a + d > 0:
            print("Tidak Tidak Ya Ya")
        elif b + c > 0:
            print("Tidak Tidak Ya Tidak")
        elif a + d > 0:
            print("Tidak Tidak Tidak Ya")
    else:
        if b + c > 0 and a + d > 0:
            print("Ya Ya Tidak Tidak")
        elif b + c > 0:
            print("Tidak Ya Tidak Tidak")
        elif a + d > 0:
            print("Ya Tidak Tidak Tidak")
    





# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()
 
 
if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
 
input = lambda: sys.stdin.readline().rstrip("\r\n")
 
# endregion
 
 
if __name__ == "__main__":
    #read()
    # sys.stdin = open("input.txt", "r")
    # sys.stdout = open("output.txt", "w")
    t = scanner()
    for i in range(t):
    	solve()
    #dmain()
 
# Comment Read()
	# fin_time = datetime.now()
# 	print("Execution time (for loop): ", (fin_time-init_time))