Template for type safe c++11 enum class flags

I am trying to create type safe C++ flags using templates. I also want to distinguish between a flag and flags (being zero, one or many flags).

The solution below works good, except for EnumFlag<T> operator | (T, T) which causes all |-operations on enums to return type EnumFlag. This breaks lots of code. Any trick fix this? In my code I do the following, however hard coding Option here is not an option. How to make this generic?

EnumFlag<typename std::enable_if<std::is_same<T, Option>::value, T>::type> operator | (T l, T r)

Changing this to...

EnumFlag<T> operator | (T l, T r)

...of cause breaks everything. I would like something like this (not compilabel code). Or any other better idea!

EnumFlag<typename std::enable_if<std::already_expanded<EnumFlag<T>>::value, T>::type> operator | (T l, T r)

Complete compilable code:

EnumFlag.h

#ifndef __classwith_flags_h_
#define __classwith_flags_h_

#include <type_traits>

enum class Option
{
    PrintHi = 1 << 0,
    PrintYo = 1 << 1,
    PrintAlot = 1 << 2
};

template <typename T>
class EnumFlag
{
public:
    using UnderlayingType = typename std::underlying_type<T>::type;

    EnumFlag(const T& flags)
        : m_flags(static_cast<UnderlayingType>(flags))
    {}

    bool operator & (T r) const
    {
        return 0 != (m_flags & static_cast<UnderlayingType>(r));
    }

    static const T NoFlag = static_cast<T>(0);

private:
    UnderlayingType  m_flags;
};
template<typename T>
EnumFlag<typename std::enable_if<std::is_same<T, Option>::value, T>::type> operator | (T l, T r)
{
    return static_cast<T>(static_cast<EnumFlag<T>::UnderlayingType>(l) | static_cast<EnumFlag<T>::UnderlayingType>(r));
}

class ClassWithFlags
{
public:
    using Options = EnumFlag < Option >;

    void doIt(const Options &options);
};

#endif

EnumFlag.cpp

#include "EnumFlag.h"

#include <iostream>

void ClassWithFlags::doIt(const Options &options)
{
    if (options & Option::PrintHi)
    {
        std::cout << "Hi" << std::endl;
    }
    if (options & Option::PrintYo)
    {
        std::cout << "Yo!" << std::endl;
    }
}

int main()
{
    ClassWithFlags classWithFlags;
    classWithFlags.doIt(Option::PrintHi | Option::PrintAlot);
}

The actual code will contain a lot more operators, however this is enough to illustrate the problem.

One less but still intrusive solution is this

template<typename T>
typename std::underlying_type<T>::type operator | (T l, T r)
{
    return (static_cast<typename std::underlying_type<T>::type>(l) | static_cast<typename std::underlying_type<T>::type>(r));
}

Not god enough, then EnumFlag(const std::underlying_type<T> &flags) must exist and I lose type safty. Also, I would like for the global operator overloads to only be created for the types actually needed. Macros is also no god because I want ta allow declaration of EnumFlags inside classes. The global overloads can not be there, hence I need two macros calls at different locations to create on EnumFlag.

The solution must be pure C++11/stl.