Lambda is deduced to std::function if template has no variadic arguments

template<typename ReturnT, typename... ParamT>
void foo(std::function<ReturnT(ParamT...)> callback)
{}

template<typename ReturnT, typename ParamT>
void bar(std::function<ReturnT(ParamT)> callback)
{}

main()
{    
    foo<int, int>([](int x){ return x; });  // no instance of function 
                                            //   template matches argument list
    bar<int, int>([](int x){ return x; });  // OK
}

The only difference between foo and bar is that foo has variadic arguments. Somehow the compiler is able to convert the lambda to a std::function in bar.

To my understanding, template type deduction doesn't consider type conversions. So shouldn't both fail?