Problem: Notes

For this problem, it is recommended to use floating point numbers, which are more accurate than double.



Input


m n x
k l y


For input, six integers m, n, x, k, l, y are given in the above input format.
These six integers correspond to those in the problem statement, repeating the horizontal m: n cut x times and the vertical k: l cut y times.
1 ≤ m, n, k, l ≤ 100, 0 ≤ x, y ≤ 40, and m, n and l, k are relatively prime.

Output

Print out the expected number of planarians that are regenerating to their original form after a few weeks in a row.
The output is acceptable with an error of 10-6 or less.

Examples

Input

1 1 1
1 1 1


Output

0.562500


Input

1 2 2
1 1 1


Output

0.490741


Input

1 2 0
3 4 0


Output

1.000000

Solution (in CPP):

Solution:
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <ctime>
#include <queue>
#include <stack>
#include <algorithm>
#include <list>
#include <vector>
#include <set>
#include <map>
#include <iostream>
#include <deque>
#include <complex>
#include <string>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <valarray>
#include <unordered_map>
#include <iterator>
#include <functional>
#include <cassert>
using namespace std;
typedef long long int ll;
typedef unsigned int uint;
typedef unsigned char uchar;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;

#define REP(i,x) for(int i=0;i<(int)(x);i++)
#define REPS(i,x) for(int i=1;i<=(int)(x);i++)
#define RREP(i,x) for(int i=((int)(x)-1);i>=0;i--)
#define RREPS(i,x) for(int i=((int)(x));i>0;i--)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();i++)
#define RFOR(i,c) for(__typeof((c).rbegin())i=(c).rbegin();i!=(c).rend();i++)
#define ALL(container) (container).begin(), (container).end()
#define RALL(container) (container).rbegin(), (container).rend()
#define SZ(container) ((int)container.size())
#define mp(a,b) make_pair(a, b)
#define pb push_back
#define eb emplace_back
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
template<class T> ostream& operator<<(ostream &os, const vector<T> &t) {
os<<"["; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"]"; return os;
}
template<class T> ostream& operator<<(ostream &os, const set<T> &t) {
os<<"{"; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"}"; return os;
}
template<class S, class T> ostream& operator<<(ostream &os, const pair<S,T> &t) { return os<<"("<<t.first<<","<<t.second<<")";}
template<class S, class T> pair<S,T> operator+(const pair<S,T> &s, const pair<S,T> &t){ return pair<S,T>(s.first+t.first, s.second+t.second);}
template<class S, class T> pair<S,T> operator-(const pair<S,T> &s, const pair<S,T> &t){ return pair<S,T>(s.first-t.first, s.second-t.second);}

const int INF = 1<<28;
const double EPS = 1e-8;
const int MOD = 1000000007;

typedef long double R;
R m, n;
int x;



template<int N>
struct Comb{
	R C[N][N];
	Comb(){
		memset(C, 0, (int)sizeof(C));	
		for(int i = 0; i < 50; i++){
			C[i][0] = C[i][i] = 1;
			for(int j = 1; j < i; j++) C[i][j] = C[i-1][j-1] + C[i-1][j];
		}
	}
	R operator()(int n, int r){
		if(n < 0 || r < 0 || r > n) return 0;
		return C[n][r];
	}
};
Comb<50> C;
R powL[51], powR[51];
R solve(R l, R r, int x){
	powL[0] = powR[0] = 1;
	REPS(i, x){
		powL[i] = powL[i-1] * l;
		powR[i] = powR[i-1] * r;
	}
	R res = 0;
	REP(i, x+1){	// i回左に行く
		R sum = C(x, i);
		REPS(j, x){	// j回目の遷移
			REP(k, j){ // 既に j - 1 回中 k 回左に遷移している
				R w = powL[k] * powR[j-1-k];
				sum -= w*(l) * C(j-1,     k) * C(x-j, i-k  );	// j回目に右に行く
			}
		}
		res += powL[i] * powR[x-i] * sum;
	}
	return res;
}

int main(int argc, char *argv[]){
	ios::sync_with_stdio(false);
	R ans = 1;
	REP(i, 2){
		cin >> m >> n >> x;
		ans *= solve(m/(n+m), n/(n+m), x);
	}
	printf("%.11f\n", (double)ans);
	return 0;
}