Problem: After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,

  * If the item was a Pink Slip, the quest is over, and you will not play any more races. 
  * Otherwise, 
    1. If a≤ v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items. 
    2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items. 



For example,

  * If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75). 
  * If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75). 
  * If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9). 
  * If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0). 



You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.

Input

The first line of input contains a single integer t (1≤ t≤ 10) — the number of test cases.

The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1≤ v≤ 0.9).

Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.

Output

For each test case, output a single line containing a single real number — the expected number of races that you must play in order to draw a Pink Slip.

Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.

Example

Input


4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048


Output


1.532000000000
1.860000000000
5.005050776521
4.260163673896

Note

For the first test case, the possible drawing sequences are: 

  * P with a probability of 0.6; 
  * CP with a probability of 0.2⋅ 0.7 = 0.14; 
  * CMP with a probability of 0.2⋅ 0.3⋅ 0.9 = 0.054; 
  * CMMP with a probability of 0.2⋅ 0.3⋅ 0.1⋅ 1 = 0.006; 
  * MP with a probability of 0.2⋅ 0.7 = 0.14; 
  * MCP with a probability of 0.2⋅ 0.3⋅ 0.9 = 0.054; 
  * MCCP with a probability of 0.2⋅ 0.3⋅ 0.1⋅ 1 = 0.006. 

So, the expected number of races is equal to 1⋅ 0.6 + 2⋅ 0.14 + 3⋅ 0.054 + 4⋅ 0.006 + 2⋅ 0.14 + 3⋅ 0.054 + 4⋅ 0.006 = 1.532.

For the second test case, the possible drawing sequences are: 

  * P with a probability of 0.4; 
  * CP with a probability of 0.4⋅ 0.6 = 0.24; 
  * CMP with a probability of 0.4⋅ 0.4⋅ 1 = 0.16; 
  * MP with a probability of 0.2⋅ 0.5 = 0.1; 
  * MCP with a probability of 0.2⋅ 0.5⋅ 1 = 0.1. 



So, the expected number of races is equal to 1⋅ 0.4 + 2⋅ 0.24 + 3⋅ 0.16 + 2⋅ 0.1 + 3⋅ 0.1 = 1.86.

Solution (in PYTHON3):

Solution:
import random
import collections
import string
import math
import copy
import os
import sys
from io import BytesIO, IOBase

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")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)


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


# mo = 998244353
mo = int(1e9+7)


def exgcd(a, b):
    if not b:
        return 1, 0
    y, x = exgcd(b, a % b)
    y -= a//b * x
    return x, y


def getinv(a, m):
    x, y = exgcd(a, m)
    return -1 if x == 1 else x % m


def comb(n, b):
    res = 1
    b = min(b, n-b)
    for i in range(b):
        res = res*(n-i)*getinv(i+1, mo) % mo
        # res %= mo
    return res % mo


def quickpower(a, n):
    res = 1
    while n:
        if n & 1:
            res = res * a % mo
        n >>= 1
        a = a*a % mo
    return res

def quickpowerm(a, n, mo):
    res = 1
    while n:
        if n & 1:
            res = res * a % mo
        n >>= 1
        a = a*a % mo
    return res


def dis(a, b):
    return abs(a[0]-b[0]) + abs(a[1]-b[1])


def getpref(x):
    if x > 1:
        return (x)*(x-1) >> 1
    else:
        return 0


def orafli(upp):
    primes = []
    marked = [False for i in range(upp+3)]
    prvs = [i for i in range(upp+3)]
    for i in range(2, upp):
        if not marked[i]:
            primes.append(i)
        for j in primes:
            if i*j >= upp:
                break
            marked[i*j] = True
            prvs[i*j] = j
            if i % j == 0:
                break
    return primes, prvs


def lower_ord(c: str) -> int:
    return ord(c)-97


def upper_ord(c: str) -> int:
    return ord(c) - 65


def read_list():
    return [int(i) for i in input().split()]


def read_int():
    s = input().split()
    if len(s) == 1:
        return int(s[0])
    else:
        return map(int, s)


def ask(s):
    print(f"? {s}", flush=True)


def answer(s):
    print(f"{s}", flush=True)

import functools

# primes, prvs = orafli(100010)

import itertools
from fractions import Fraction


# A = list(map(list,(zip(*A))))

def solve():
    cache = {}
    # n = read_int()
    c, m, p, v = [float(i) for i in input().split()]
    v2 = v / 2
    def dfs(cc, mm, pp, dep) -> float:
        # if (cc, mm, pp) in cache:
            # return dep + cache.get((cc,mm,pp))
        # else:
        res = pp * dep
        if cc > 1e-10:
            if cc > v:
                if mm < 1e-10:
                    res += cc * dfs(cc-v, 0, pp+v, dep+1)
                else:
                    res += cc * dfs(cc-v, mm+v2, pp+v2, dep+1)
            else:
                if mm < 1e-10:
                    res += cc * dfs(0, 0, 1, dep+1)
                else:
                    res += cc * dfs(0, mm+cc/2, pp+cc/2, dep+1)
        if mm > 1e-10:
            if mm > v:
                if cc < 1e-10:
                    res += mm * dfs(0, mm-v, pp+v, dep+1)
                else:
                    res += mm * dfs(cc+v2, mm-v, pp+v2, dep+1)
            else:
                if cc < 1e-10:
                    res += mm * dfs(0, 0, 1, dep+1)
                else:
                    res += mm * dfs(cc+mm/2, 0, pp+mm/2, dep+1)
        
        # cache[(cc,mm,pp)] = res
        return res
    rr = dfs(c,m,p,1)
    print(rr)

    
# fi = open('C:\\cppHeaders\\CF2020.12.17\\test.data', 'r')
# def input(): return fi.readline().rstrip("\r\n")
# primes, prv = orafli(10001)
# solve()
T = int(input())
# t = 1



for ti in range(T):
    solve()
    # except:
        # traceback.print_exc()