1 //#include<bits/stdc++.h>

  2 #include<time.h>

  3 #include <set>

  4 #include <map>

  5 #include <stack>

  6 #include <cmath>

  7 #include <queue>

  8 #include <cstdio>

  9 #include <string>

 10 #include <vector>

 11 #include <cstring>

 12 #include <utility>

 13 #include <cstring>

 14 #include <iostream>

 15 #include <algorithm>

 16 #include <list>

 17 using namespace std;

 18 #define eps 1e-10

 19 #define PI acos(-1.0)

 20 #define lowbit(x) ((x)&(-x))

 21 #define zero(x) (((x)>0?(x):-(x))<eps)

 22 #define mem(s,n) memset(s,n,sizeof s);

 23 #define ios {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}

 24 typedef long long ll;

 25 typedef unsigned long long ull;

 26 const int maxn=1e3+5;

 27 const int Inf=0x7f7f7f7f;

 28 const ll Mod=999911659;

 29 const int N=3e3+5;

 30 bool isPowerOfTwo(int n) { return n > 0 && (n & (n - 1)) == 0; }//判断一个数是不是 2 的正整数次幂

 31 int modPowerOfTwo(int x, int mod) { return x & (mod - 1); }//对 2 的非负整数次幂取模

 32 int getBit(int a, int b) { return (a >> b) & 1; }// 获取 a 的第 b 位，最低位编号为 0

 33 int Max(int a, int b) { return b & ((a - b) >> 31) | a & (~(a - b) >> 31); }// 如果 a>=b,(a-b)>>31 为 0，否则为 -1

 34 int Min(int a, int b) { return a & ((a - b) >> 31) | b & (~(a - b) >> 31); }

 35 ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}

 36 ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}

 37 int Abs(int n) {

 38   return (n ^ (n >> 31)) - (n >> 31);

 39   /* n>>31 取得 n 的符号，若 n 为正数，n>>31 等于 0，若 n 为负数，n>>31 等于 -1

 40      若 n 为正数 n^0=n, 数不变，若 n 为负数有 n^(-1)

 41      需要计算 n 和 -1 的补码，然后进行异或运算，

 42      结果 n 变号并且为 n 的绝对值减 1，再减去 -1 就是绝对值 */

 43 }

 44 ll binpow(ll a, ll b,ll c) {

 45   ll res = 1;

 46   while (b > 0) {

 47     if (b & 1) res = res * a%c;

 48     a = a * a%c;

 49     b >>= 1;

 50   }

 51   return res%c;

 52 }

 53 void extend_gcd(ll a,ll b,ll &x,ll &y)

 54 {

 55     if(b==0) {

 56         x=1,y=0;

 57         return;

 58     }

 59     extend_gcd(b,a%b,x,y);

 60     ll tmp=x;

 61     x=y;

 62     y=tmp-(a/b)*y;

 63 }

 64 ll mod_inverse(ll a,ll m)

 65 {

 66     ll x,y;

 67     extend_gcd(a,m,x,y);

 68     return (m+x%m)%m;

 69 }

 70 ll eulor(ll x)

 71 {

 72    ll cnt=x;

 73    ll ma=sqrt(x);

 74    for(int i=2;i<=ma;i++)

 75    {

 76     if(x%i==0) cnt=cnt/i*(i-1);

 77     while(x%i==0) x/=i;

 78    }

 79    if(x>1) cnt=cnt/x*(x-1);

 80    return cnt;

 81 }

 82 double br,tr,H,v;

 83 double getv(double h)

 84 {

 85     double rr=(tr-br)*h/H+br;

 86     return PI*h*(br*br+br*rr+rr*rr)/3;

 87 }

 88 int main()

 89 {

 90     int t;

 91     cin>>t;

 92     while(t--)

 93     {

 94         cin>>br>>tr>>H>>v;

 95        double l=0,r=H,mid;

 96        while(r-l>eps)

 97        {

 98            mid=(l+r)/2;

 99            if(fabs(getv(mid)-v)<=eps) break;

100            else if(fabs((getv(mid)))>v) r=mid;

101            else l=mid;

102        }

103        printf("%.6f\n",mid);

104     }

105     return 0;

106 }