Why specialization based on enable_if is not picked up by compiler

I'd like to specialize class for some type of classes, for example based on std::is_arithmetic. Although compiler doesn't "see" my specialization based on "enable_if" and chooses principle/main template. Could you please help me with this... Below is snippet of code and output after compiling with g++ 4.8

#include < iostream >  
#include < type_traits >  
#include < string >  

template < typename T1, typename T2 = void >  
struct TestT  
{  
    static const bool is_int = false;  
    static const bool is_str = false;  
};

template < typename T>  
struct TestT < T,  
       std::enable_if< std::is_arithmetic<t>::value, T >::type >  
{  
    static const bool is_int = false;  
    static const bool is_str = false;  
};  

template < typename T>
struct TestT < std::string, T >  
{  
    static const bool is_int = false;  
    static const bool is_str = true;  
};  

class enum TestE  
{  
    Last  
};

int main(int argc, char* argv[])  
{
    std::cout << "Enum is_int: " << TestT<TestE>::is_int  
              << ", is_str: " << TestT<TestE>::is_str << std::endl;  
    std::cout << "string is_int: " << TestT<std::string>::is_int  
              << ", is_str: " << TestT<std::string>::is_str << std::endl;  
    std::cout << "int is_int: " << TestT<int>::is_int  
              << ", is_str: " << TestT<int>::is_str << std::endl;  
    return 0;
}  

Output of above is:

Enum is_int: 0, is_str: 0 // Expected
string is_int: 0, is_str: 1 // Expected
int is_int: 0, is_str: 0 // Not expected

I really would appreciate for any help, and thank you in advance