Problem: There are two (6-sided) dice: a red die and a blue die. When a red die is rolled, it shows i with probability p_i percents, and when a blue die is rolled, it shows j with probability q_j percents.

Petr and tourist are playing the following game. Both players know the probabilistic distributions of the two dice. First, Petr chooses a die in his will (without showing it to tourist), rolls it, and tells tourist the number it shows. Then, tourist guesses the color of the chosen die. If he guesses the color correctly, tourist wins. Otherwise Petr wins.

If both players play optimally, what is the probability that tourist wins the game?

Constraints

* 0 ≤ p_i, q_i ≤ 100
* p_1 + ... + p_6 = q_1 + ... + q_6 = 100
* All values in the input are integers.

Input

The input is given from Standard Input in the following format:


p_1 p_2 p_3 p_4 p_5 p_6
q_1 q_2 q_3 q_4 q_5 q_6


Output

Print the probability that tourist wins. The absolute error or the relative error must be at most 10^{-9}.

Examples

Input

25 25 25 25 0 0
0 0 0 0 50 50


Output

1.000000000000


Input

10 20 20 10 20 20
20 20 20 10 10 20


Output

0.550000000000

Solution (in CPP):

Solution:
/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author RiaD
 */

#include <iostream>
#include <fstream>

#include <iostream>


#include <iterator>


#include <string>
#include <stdexcept>

#ifndef SPCPPL_ASSERT
	#ifdef SPCPPL_DEBUG
		#define SPCPPL_ASSERT(condition) \
		if(!(condition)) { \
			throw std::runtime_error(std::string() + #condition + " in line " + std::to_string(__LINE__) + " in " + __PRETTY_FUNCTION__); \
		}
	#else
		#define SPCPPL_ASSERT(condition)
	#endif
#endif


/**
* Support decrementing and multi-passing, but not declared bidirectional(or even forward) because
* it's reference type is not a reference.
*
* It doesn't return reference because
* 1. Anyway it'll not satisfy requirement [forward.iterators]/6
*   If a and b are both dereferenceable, then a == b if and only if *a and
*   b are bound to the same object.
* 2. It'll not work with reverse_iterator that returns operator * of temporary which is temporary for this iterator
*
* Note, reverse_iterator is not guaranteed to work  now too since it works only with bidirectional iterators,
* but it's seems to work at least on my implementation.
*
* It's not really useful anywhere except iterating anyway.
*/
template <typename T>
class IntegerIterator: public std::iterator<std::input_iterator_tag, T, std::ptrdiff_t, T*, T> {
public:
	explicit IntegerIterator(T value): value(value) {

	}

	IntegerIterator& operator++() {
		++value;
		return *this;
	}

	IntegerIterator operator++(int) {
		IntegerIterator copy = *this;
		++value;
		return copy;
	}

	IntegerIterator& operator--() {
		--value;
		return *this;
	}

	IntegerIterator operator--(int) {
		IntegerIterator copy = *this;
		--value;
		return copy;
	}

	T operator*() const {
		return value;
	}

	bool operator==(IntegerIterator rhs) const {
		return value == rhs.value;
	}

	bool operator!=(IntegerIterator rhs) const {
		return !(*this == rhs);
	}

private:
	T value;
};

template <typename T>
class IntegerRange {
public:
	IntegerRange(T begin, T end): begin_(begin), end_(end) {
		SPCPPL_ASSERT(begin <= end);
	}

	IntegerIterator<T> begin() const {
		return IntegerIterator<T>(begin_);
	}

	IntegerIterator<T> end() const {
		return IntegerIterator<T>(end_);
	}

private:
	T begin_;
	T end_;
};

template <typename T>
class ReversedIntegerRange {
	typedef std::reverse_iterator<IntegerIterator<T>> IteratorType;
public:
	ReversedIntegerRange(T begin, T end): begin_(begin), end_(end) {
		SPCPPL_ASSERT(begin >= end);
	}

	IteratorType begin() const {
		return IteratorType(IntegerIterator<T>(begin_));
	}

	IteratorType end() const {
		return IteratorType(IntegerIterator<T>(end_));
	}

private:
	T begin_;
	T end_;
};

template <typename T>
IntegerRange<T> range(T to) {
	return IntegerRange<T>(0, to);
}

template <typename T>
IntegerRange<T> range(T from, T to) {
	return IntegerRange<T>(from, to);
}

template <typename T>
IntegerRange<T> inclusiveRange(T to) {
	return IntegerRange<T>(0, to + 1);
}

template <typename T>
IntegerRange<T> inclusiveRange(T from, T to) {
	return IntegerRange<T>(from, to + 1);
}

template <typename T>
ReversedIntegerRange<T> downrange(T from) {
	return ReversedIntegerRange<T>(from, 0);
}

template <typename T>
ReversedIntegerRange<T> downrange(T from, T to) {
	return ReversedIntegerRange<T>(from, to);
}

template <typename T>
ReversedIntegerRange<T> inclusiveDownrange(T from) {
	return ReversedIntegerRange<T>(from + 1, 0);
}

template <typename T>
ReversedIntegerRange<T> inclusiveDownrange(T from, T to) {
	return ReversedIntegerRange<T>(from + 1, to);
}

#include <vector>

using namespace std;

class TaskD {
public:
	void solve(std::istream& in, std::ostream& out) {
		vector<int> a(6), b(6);
		for (int i: range(6)) {
			in >> a[i];
		}
		for (int i: range(6)) {
			in >> b[i];
		}

		auto solve = [&](double p) {
			auto q = 1 - p;
			double ans = 0;
			for (int i: range(6)) {
				ans += max(p * a[i], q * b[i]);
			}
			return ans;
		};

		//cerr << solve(0.5) << endl;

		double mnAns = 1000000;

		for (int i: range(3000)) {
			for (int j: range(i + 1)) {
				mnAns = min(mnAns, solve(j * 1.0 / i));
			}
		}
		out << mnAns / 100.0 << endl;
	}
};


int main() {
	std::ios_base::sync_with_stdio(false);
	TaskD solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	in.tie(nullptr);
	out << std::fixed;
	out.precision(20);
	solver.solve(in, out);
	return 0;
}