What is the computational complexity of this SAT Variant Given a 3SAT problem. The question being: 'This Problem has exactly K Solutions'?

Now, lets say K=1 (without loss of generality).


If the problem has a exactly 1 solution and the answer is True. So, the partial certificate is that 1 solution. the certificate is partial as it shows there is a solution but without guarantee of exactly 1.  
If the answer is False. Then either it has 0 or more than 1 solution.

(a) In case of more than 1 solution, the certificate is any of the 2 or more solutions. This is a complete certificate.

(b) In case of 0 solution, the problem is same as asking does the given SAT have a solution. The certificate being same in 1.


Thus, the problem seems to require a certificate from both NP and co-NP versions. It seems to mean problem lies in NP and co-NP. But then it also is NP-hard. So doesn't that imply NP=co-NP?! What am I missing.
The class ${\bf C_=P}$ is defined to be the set of languagges $L$ such that there exist functions $f \in {\bf \# P}$,$t\in {\bf FP} $ and for all $x, x\in L$ if and only if $f(x)=t(x)$.  Exact-SAT is the standard complete language for ${\bf C_=P}$. Exact-SAT = $\{F, k\} | F$ is a Boolean formula with exactly $k$ satisfying truth assignments.  It is known that ${\bf CoNP} \subseteq {\bf US} \subseteq {\bf C_=P}$  See the paper "On the Power of Deterministic Reductions to ${\bf C_=P}$ " by Frederic Green.