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:
# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations  # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,

# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect

# If the element is already present in the list,
# the right most position where element has to be inserted is returned

# ==============================================================================================
# fast I/O region

BUFSIZE = 8192
from sys import stderr


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)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType


def iterative(f, stack=[]):
    def wrapped_func(*args, **kwargs):
        if stack: return f(*args, **kwargs)
        to = f(*args, **kwargs)
        while True:
            if type(to) is GeneratorType:
                stack.append(to)
                to = next(to)
                continue
            stack.pop()
            if not stack: break
            to = stack[-1].send(to)
        return to

    return wrapped_func


#### END ITERATE RECURSION ####

# ===============================================================================================
# some shortcuts

mod = 1000000007


def inp(): return sys.stdin.readline().rstrip("\r\n")  # for fast input


def out(var): sys.stdout.write(str(var))  # for fast output, always take string


def lis(): return list(map(int, inp().split()))


def stringlis(): return list(map(str, inp().split()))


def sep(): return map(int, inp().split())


def strsep(): return map(str, inp().split())


def fsep(): return map(float, inp().split())


def nextline(): out("\n")  # as stdout.write always print sring.


def testcase(t):
    for p in range(t):
        solve()


def pow(x, y, p):
    res = 1  # Initialize result
    x = x % p  # Update x if it is more , than or equal to p
    if (x == 0):
        return 0
    while (y > 0):
        if ((y & 1) == 1):  # If y is odd, multiply, x with result
            res = (res * x) % p

        y = y >> 1  # y = y/2
        x = (x * x) % p
    return res


from functools import reduce


def factors(n):
    return set(reduce(list.__add__,
                      ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))


def gcd(a, b):
    if a == b: return a
    while b > 0: a, b = b, a % b
    return a


# discrete binary search
# minimise:
# def search():
#     l = 0
#     r = 10 ** 15
#
#     for i in range(200):
#         if isvalid(l):
#             return l
#         if l == r:
#             return l
#         m = (l + r) // 2
#         if isvalid(m) and not isvalid(m - 1):
#             return m
#         if isvalid(m):
#             r = m + 1
#         else:
#             l = m
#     return m

# maximise:
# def search():
#     l = 0
#     r = 10 ** 15
#
#     for i in range(200):
#         # print(l,r)
#         if isvalid(r):
#             return r
#         if l == r:
#             return l
#         m = (l + r) // 2
#         if isvalid(m) and not isvalid(m + 1):
#             return m
#         if isvalid(m):
#             l = m
#         else:
#             r = m - 1
#     return m


##to find factorial and ncr
# N=100000
# mod = 10**9 +7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, N + 1):
#     fac.append((fac[-1] * i) % mod)
#     inv.append(mod - (inv[mod % i] * (mod // i) % mod))
#     finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
#     if n < r:
#         return 0
#     else:
#         return fac[n] * (finv[r] * finv[n - r] % mod) % mod


##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
#     dp[pos]=0
#     l=max(1,k+pos-n-1)
#     for j in range(min(pos,k),l-1,-1):
#         dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])

def prefix_sum(ar):  # [1,2,3,4]->[1,3,6,10]
    return list(accumulate(ar))


def suffix_sum(ar):  # [1,2,3,4]->[10,9,7,4]
    return list(accumulate(ar[::-1]))[::-1]


def N():
    return int(inp())


# =========================================================================================
from collections import defaultdict



def numberOfSetBits(i):
    i = i - ((i >> 1) & 0x55555555)
    i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
    return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24



def solve():
    a,b,c,d=sep()
    ar=[0]*4
    if (a+b)%2==1:
        if a>0 or d>0:
            ar[0]=1
        if b>0 or c>0:
            ar[1]=1
    else:
        if a>0 or d>0:
            ar[3]=1
        if b>0 or c>0:
            ar[2]=1
    for i in ar:
        if i==1:
            print("Ya",end=" ")
        else:
            print("Tidak",end=" ")
    print()

















# solve()
testcase(int(inp()))