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 CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define int               long long int
#define pb                push_back
#define ppb               pop_back
#define pf                push_front
#define ppf               pop_front
#define all(x)            (x).begin(),(x).end()
#define uniq(v)           (v).erase(unique(all(v)),(v).end())
#define sz(x)             (int)((x).size())
#define fr                first
#define sc                second
#define pii               pair<int,int>
#define rep(i,a,b)        for(int i=a;i<b;i++)
#define mem1(a)           memset(a,-1,sizeof(a))
#define mem0(a)           memset(a,0,sizeof(a))
#define ppc               __builtin_popcount
#define ppcll             __builtin_popcountll

template<typename T1,typename T2>istream& operator>>(istream& in,pair<T1,T2> &a){in>>a.fr>>a.sc;return in;}
template<typename T1,typename T2>ostream& operator<<(ostream& out,pair<T1,T2> a){out<<a.fr<<a.sc;return out;}
template<typename T,typename T1>T amax(T &a,T1 b){if(b>a)a=b;return a;}
template<typename T,typename T1>T amin(T &a,T1 b){if(b<a)a=b;return a;}
const int N5=1e5+7,N5_2=2e5+7,N5_3=3e5+7,N6=1e6+7,N7=1e7+7,mod=1e9+7;
// int prime[N6];
long double ex=0;
long double scale=1e7;

bool comp(pair<int,int> a, pair<int,int> b){
    return a.second<b.second?1:0;
}
 
int pw(int n, int p,int M=mod){
    int val=1;
    while(p>0){
        if(p & 1){
            val= (val*n)%M;
        }
        n=(n*n)%M;
        p>>=1;
    }
    return val;
}
 
// void SieveofEratosthenes(int n=N6){
//     memset(prime,true, sizeof(prime));
//     rep(i,2,n+1){
//         if(prime[i]){
//             for(int j=i*i;j<=n;j+=i){
//                 prime[j]=false;
//             }
//         }
//     }
//     return;
// }

/*
1010
0000
1001
0000
*/

/*
57

6 6 6 6 6 6 6 5 5 5

*/

void rec(int c,int m, int p,int v,int count,long double prob){
    ex+=count*(prob==0?p/scale:prob*(p/scale));
    // cout<<"p = "<<count*(prob==0?p:prob*p)<<endl;
    if(c>v){
        // rep(i,0,count)  cout<<" ";
        // cout<<"c";
        if(m>0)
            rec(c-v,m+v/(2.0),p+v/(2.0),v,count+1,prob==0?c/scale:prob*(c/scale));
        else
            rec(c-v,0,p+v,v,count+1,prob==0?c/scale:prob*(c/scale));
    }
    else if(c>0){
        // rep(i,0,count)  cout<<" ";
        // cout<<"c";
        if(m>0)
            rec(0,m+c/(2.0),p+c/(2.0),v,count+1,prob==0?c/scale:prob*(c/scale));
        else{
            rec(0,0,p+c,v,count+1,prob==0?c/scale:prob*(c/scale));
        }
    }
    if(m>v){
        // rep(i,0,count)  cout<<" ";
        // cout<<"m";
        if(c>0)
            rec(c+v/(2.0),m-v,p+v/(2.0),v,count+1,prob==0?m/scale:prob*(m/scale));
        else{
            rec(0,m-v,p+v,v,count+1,prob==0?m/scale:prob*(m/scale));
        }
    }
    else if(m>0){
        // rep(i,0,count)  cout<<" ";
        // cout<<"m";
        if(c>0)
            rec(c+m/(2.0),0,p+m/(2.0),v,count+1,prob==0?m/scale:prob*(m/scale));
        else
            rec(0,0,p+m,v,count+1,prob==0?m/scale:prob*(m/scale));
    }
    return;
}


void solve(){
    long double cd,md,pd,vd;
    cin>>cd>>md>>pd>>vd;
    ex=0;
    int c = round(scale*cd);
    int m = round(scale*md);
    int p = round(scale*pd);
    int v = round(scale*vd);
    rec(c,m,p,v,1,0);
    cout<<ex<<endl;
    return;

}

signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    #ifndef ONLINE_JUDGE
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    #endif
    #ifdef SIEVE
        sieve();
    #endif
    #ifdef NCR
        init();
    #endif
    cout<<setprecision(8);
    int t=1;
    cin>>t;
    while(t--) solve();
    return 0;
}