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 JAVA):

Solution:
import java.beans.Visibility;
import java.io.*;
import java.util.*;
import java.math.BigInteger;

/**
 __                  __
 ( _)                ( _)
 / / \\              / /\_\_
 / /   \\            / / | \ \
 / /     \\          / /  |\ \ \
 /  /   ,  \ ,       / /   /|  \ \
 /  /    |\_ /|      / /   / \   \_\
 /  /  |\/ _ '_| \   / /   /   \    \\
 |  /   |/  0 \0\    / |    |    \    \\
 |    |\|      \_\_ /  /    |     \    \\
 |  | |/    \.\ o\o)  /      \     |    \\
 \    |     /\\`v-v  /        |    |     \\
 | \/    /_| \\_|  /         |    | \    \\
 | |    /__/_ `-` /   _____  |    |  \    \\
 \|    [__]  \_/  |_________  \   |   \    ()
 /    [___] (    \         \  |\ |   |   //
 |    [___]                  |\| \|   /  |/
 /|    [____]                  \  |/\ / / ||
 (  \   [____ /     ) _\      \  \    \| | ||
 \  \  [_____|    / /     __/    \   / / //
 |   \ [_____/   / /        \    |   \/ //
 |   /  '----|   /=\____   _/    |   / //
 __ /  /        |  /   ___/  _/\    \  | ||
 (/-(/-\)       /   \  (/\/\)/  |    /  | /
 (/\/\)           /   /   //
 _________/   /    /
 \____________/    (


 @author NTUDragons-Reborn
 */

public class C{
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }
    // main solver
    static class Task{

        double eps= 0.00000001;
        static final int MAXN = 100005;
        static final int MOD= 1000000007;

        // stores smallest prime factor for every number
        static int spf[] = new int[MAXN];

        static boolean[] prime;

        Map<Integer,Set<Integer>> dp= new HashMap<>();

        // Calculating SPF (Smallest Prime Factor) for every
        // number till MAXN.
        // Time Complexity : O(nloglogn)
        public void sieve()
        {
            spf[1] = 1;
            for (int i=2; i<MAXN; i++)

                // marking smallest prime factor for every
                // number to be itself.
                spf[i] = i;

            // separately marking spf for every even
            // number as 2
            for (int i=4; i<MAXN; i+=2)
                spf[i] = 2;

            for (int i=3; i*i<MAXN; i++)
            {
                // checking if i is prime
                if (spf[i] == i)
                {
                    // marking SPF for all numbers divisible by i
                    for (int j=i*i; j<MAXN; j+=i)

                        // marking spf[j] if it is not
                        // previously marked
                        if (spf[j]==j)
                            spf[j] = i;
                }
            }
        }

        void sieveOfEratosthenes(int n)
        {
            // Create a boolean array
            // "prime[0..n]" and
            // initialize all entries
            // it as true. A value in
            // prime[i] will finally be
            // false if i is Not a
            // prime, else true.
            prime= new boolean[n + 1];
            for (int i = 0; i <= n; i++)
                prime[i] = true;
    
            for (int p = 2; p * p <= n; p++)
            {
                // If prime[p] is not changed, then it is a
                // prime
                if (prime[p] == true)
                {
                    // Update all multiples of p
                    for (int i = p * p; i <= n; i += p)
                        prime[i] = false;
                }
            }
        }

        // A O(log n) function returning primefactorization
        // by dividing by smallest prime factor at every step
        public Set<Integer> getFactorization(int x)
        {
            if(dp.containsKey(x)) return dp.get(x);
            Set<Integer> ret = new HashSet<>();
            while (x != 1)
            {
                if(spf[x]!=2) ret.add(spf[x]);
                x = x / spf[x];
            }
            dp.put(x,ret);
            return ret;
        }
        // function to find first index >= x
        public int lowerIndex(List<Integer> arr, int n, int x)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr.get(mid) >= x)
                    h = mid - 1;
                else
                    l = mid + 1;
            }
            return l;
        }
        public int lowerIndex(int[] arr, int n, int x)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr[mid] >= x)
                    h = mid - 1;
                else
                    l = mid + 1;
            }
            return l;
        }

        // function to find last index <= y
        public int upperIndex(List<Integer> arr, int n, int y)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr.get(mid) <= y)
                    l = mid + 1;
                else
                    h = mid - 1;
            }
            return h;
        }
        public int upperIndex(int[] arr, int n, int y)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr[mid] <= y)
                    l = mid + 1;
                else
                    h = mid - 1;
            }
            return h;
        }

        // function to count elements within given range
        public int countInRange(List<Integer> arr, int n, int x, int y)
        {
            // initialize result
            int count = 0;
            count = upperIndex(arr, n, y) -
                    lowerIndex(arr, n, x) + 1;
            return count;
        }

        public int _gcd(int a,  int  b)
        {

            if(b == 0) {
                return a;
            }
            else {
                return _gcd(b, a % b);
            }
        }

        public int add(int a, int b){
            a+=b;
            if(a>=MOD) a-=MOD;
            else if(a<0) a+=MOD;
            return a;
        }

        public int mul(int a, int b){
            long res= (long)a*(long)b;
            return (int)(res%MOD);
        }

        public int power(int a, int b) {
            int ans=1;
            while(b>0){
                if((b&1)!=0) ans= mul(ans,a);
                b>>=1;
                a= mul(a,a);
            }
            return ans;
        }

        int[] fact= new int[MAXN];
        int[] inv= new int[MAXN];

        public int Ckn(int n, int k){
            if(k<0 || n<0) return 0;
            return mul(mul(fact[n],inv[k]),inv[n-k]);
        }

        public int inverse(int a){
            return power(a,MOD-2);
        }

        public void preprocess() {
            fact[0]=1;
            for(int i=1;i<MAXN;i++) fact[i]= mul(fact[i-1],i);
            inv[MAXN-1]= inverse(fact[MAXN-1]);
            for(int i=MAXN-2;i>=0;i--){
                inv[i]= mul(inv[i+1],i+1);
            }
        }

        /**
         * return VALUE of lower bound for unsorted array
         */
        public int lowerBoundNormalArray(int[] arr, int x){
            TreeSet<Integer> set= new TreeSet<>();
            for(int num: arr) set.add(num);
            return set.lower(x);
        }
        /**
         * return VALUE of upper bound for unsorted array
         */
        public int upperBoundNormalArray(int[] arr, int x){
            TreeSet<Integer> set= new TreeSet<>();
            for(int num: arr) set.add(num);
            return set.higher(x);
        }

        public void debugArr(int[] arr){
            for(int i: arr) out.print(i+" ");
            out.println();
        }
        
        InputReader in; PrintWriter out;
        
        public void solve(InputReader in, PrintWriter out) {
            this.in=in; this.out=out;
            int t= in.nextInt();
            while(t-->0){
                solveC();
            }
        }

        public void solveB(){
            long a= in.nextLong(), b= in.nextLong();
            long gcd=  Math.abs(a-b);
            if(gcd==0){
                out.println("0 0");
                return;
            }
            long temp= a/gcd;
            if(a%gcd!=0) temp++;
            long top= temp*gcd;
            long res= Math.min(top-a,Math.min(a,b));
            out.println(gcd+" "+res);
            
        }

        double c,m,p,v;
        public void solveC(){
            res=0;
            c= in.nextDouble(); m= in.nextDouble(); p= in.nextDouble(); v=in.nextDouble();
            double[] state= new double[]{c,m,p};
            go(0,state,(double)1.0);
            out.println(res);

        }
        double res=0;
        double EPS= 0.000001;

        public void go(int steps, double[] state, double cur){
            //end here-choose P
            res+= (double)(steps+1)*(cur*state[2]);
            // out.println((steps+1)+" "+(cur*state[2]));
            // out.println("state: "+state[2]);
            if(Math.abs(state[2]-1)<EPS) return;
            if(state[0]<EPS) state[0]=-1;
            if(state[1]<EPS) state[1]=-1;

            int num=0;
            if(state[0]>0) num++;
            if(state[1]>0) num++;
            if(state[2]>0) num++;
            //go c
            double[] state1= new double[]{state[0],state[1],state[2]};
            double[] state2= new double[]{state[0],state[1],state[2]};
            if(state1[0]>0){
                num--;
                double newCur= cur*state1[0];
                if(state1[0]>v){
                    state1[0]-=v; 
                    if(state1[1]>0) state1[1]+= v/num;
                    if(state1[2]>0) state1[2]+= v/num;
                }
                else{
                    if(state1[1]>0) state1[1]+= state1[0]/num;
                    if(state1[2]>0) state1[2]+= state1[0]/num;
                    state1[0]=-1;
                }
                go(steps+1,state1,newCur);
            }
            num=0;
            if(state[0]>0) num++;
            if(state[1]>0) num++;
            if(state[2]>0) num++;
            //go m
            if(state2[1]>0){
                num--;
                double newCur= cur*state2[1];
                if(state2[1]>v){
                    state2[1]-=v; 
                    if(state2[0]>0) state2[0]+= v/num;
                    if(state2[2]>0) state2[2]+= v/num;
                }
                else{
                    if(state2[0]>0) state2[0]+= state2[1]/num;
                    if(state2[2]>0) state2[2]+= state2[1]/num;
                    state2[1]=-1;
                }
                go(steps+1,state2,newCur);
            }

        }

    }

    static class Venice{
        public Map<Long,Long> m= new HashMap<>();
        public long base=0;
        public long totalValue=0;
        private int M= 1000000007;

        private long addMod(long a, long b){
            a+=b;
            if(a>=M) a-=M;
            return a;
        }

        public void reset(){
            m= new HashMap<>();
            base=0;
            totalValue=0;
        }
        public void update(long add){
            base= base+ add;
        }
        public void add(long key, long val){
            long newKey= key-base;
            m.put(newKey, addMod(m.getOrDefault(newKey,(long)0),val));
        }
    }


    static class Tuple implements Comparable<Tuple>{
        int x, y, z;
        public Tuple(int x, int y, int z){
            this.x= x;
            this.y= y;
            this.z=z;
        }
        @Override
        public int compareTo(Tuple o){
            int val1= this.x-this.y;
            int val2= o.x-o.y;
            return val1-val2;
        }
    }

    static class Pair implements Comparable<Pair>{
        public int x;
        public int y;
        public Pair(int x, int y){
            this.x= x;
            this.y= y;
        }

        @Override
        public int compareTo(Pair o) {
            int val1= this.x-this.y;
            int val2= o.x-o.y;
            return val1-val2;
        }
    }

    // public static class compareL implements Comparator<Tuple>{
    //     @Override
    //     public int compare(Tuple t1, Tuple t2) {
    //         return t2.l - t1.l;
    //     }
    // }

    // fast input reader class;
    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                String line = null;
                try {
                    line = br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (line == null) {
                    return null;
                }
                st = new StringTokenizer(line);
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        public double nextDouble(){
            return Double.parseDouble(nextToken());
        }
        public long nextLong(){
            return Long.parseLong(nextToken());
        }
        public int[] nextIntArr(int n){
            int[] arr= new int[n];
            for(int i=0;i<n;i++) arr[i]= nextInt();
            return arr;
        }
        public long[] nextLongArr(int n){
            long[] arr= new long[n];
            for(int i=0;i<n;i++) arr[i]= nextLong();
            return arr;
        }
        public List<Integer> nextIntList(int n){
            List<Integer> arr= new ArrayList<>();
            for(int i=0;i<n;i++) arr.add(nextInt());
            return arr;
        }
        public int[][] nextIntMatArr(int n, int m){
            int[][] mat= new int[n][m];
            for(int i=0;i<n;i++) for(int j=0;j<m;j++) mat[i][j]= nextInt();
            return mat;
        }
        public List<List<Integer>> nextIntMatList(int n, int m){
            List<List<Integer>> mat= new ArrayList<>();
            for(int i=0;i<n;i++){
                List<Integer> temp= new ArrayList<>();
                for(int j=0;j<m;j++) temp.add(nextInt());
                mat.add(temp);
            }
            return mat;
        }
        public char[] nextStringCharArr(){
            return nextToken().toCharArray();
        }
    }
}