Re: template friend for operator- How to

I have a problem with a friend declaration of the dyadic operator- IFF there is a monadic operator-, as well.

Note that the friend declaration below is accepted if and only if the definition of the monadic operator
is removed.

Many thanks for a hint,
Helmut

template <typename INT,INT P> class Zp;

template <typename INT,INT P>
Zp<INT,P> operator-(const Zp<INT,P>& a, const Zp<INT,P>& b);

template <typename INT,INT P>

class Zp {
public:
   static const INT p = P;
private:

   INT val;

public:

   Zp() : val(0) {}
   Zp( INT x ) : val(x%p) { if (x < 0 ) x+= p; }

   Zp operator-() const { return Zp(p-val); } // if this is deleted, it works <<<<<
   friend Zp<INT,P> operator- <>(const Zp<INT,P>& a, const Zp<INT,P>& b);

/* gcc-4.9.0 says
Quest_Templ.C:22:28: error: declaration of 'operator-' as non-function
    friend Zp<INT,P> operator- <>(const Zp<INT,P>& a, const Zp<INT,P>& b);
                             ^
Quest_Templ.C:22:28: error: expected ';' at end of member declaration
Quest_Templ.C:22:30: error: expected unqualified-id before '<' token
    friend Zp<INT,P> operator- <>(const Zp<INT,P>& a, const Zp<INT,P>& b);
*/
};