] \E[ X_{j_1} \cdots X_{j_{q}} ]  \big) & \text{if}\ p,  q \mbox{ both are odd,}\\\\
            0 &  \text{otherwise}.       
              \end{array}\right.                  
    \end{aligned}$$ Now we calculate right hand side of (\[eqn:lim\_cov\_SC\]). Depending on values of $p,q$, following two subcases arise.\
\
**subcase (i)** **$p,q$ both are even:** First recall, the typical term of $I_{2r}$ is $$((d_1, d_2,\ldots, d_{2r}, i_{2r+1}, \ldots, i_{p}), (d_1,d_2,\ldots, d_{2r}, j_{2r+1}, \ldots, j_{q})).$$ For such an element of $I_{2r}$, the number of free entries in $I_{2r}$ will be maximum, if following conditions hold

1.  $ \{d_1, d_2, \ldots, d_{2r}\} \cap \{i_{2r+1},\ldots,i_{p}\}\cap \{j_{2r+1},\ldots,j_{q}\}=\emptyset$,

2.  $(i_{2r+1}, \ldots, i_{p})$ and $(j_{2r+1}, \ldots, j_{q})$ are [*opposite sign pair matched*]{}.

Due to the above consideration, the constraints, $\sum_{t=1}^{2r}\epsilon_t d_t +\sum_{t=2r+1}^{p}\epsilon_t i_t =0\; \mbox{(mod n})$ and $\sum_{t=1}^{2r}\epsilon_t d_t  + \sum_{t=2r+1}^{q}\epsilon_t j_t =0\; \mbox{(mod n})$ will change into one constraint $$\label{eqn:constraint_d_even}
\sum_{t=1}^{2r}\epsilon_t d_t =0\; \mbox{(mod n}).$$ Now first we consider $r\geq 2$, later we shall deal $r=1$ case. Note that for $r=2,3, \ldots, \min\{p,q\}$, if we assume each entries of $ \{d_1, d_2, \ldots, d_{2r}\}$ are distinct, then cardinality of $I_{2r}$ will be of the order $O(n^{2r-1 +\frac{p-2r}{2} +\frac{q-2r}{2} })= O(n^{\frac{p+q}{2} -1}),$ where $(-1)$ is arising due to (\[eqn:constraint\_d\_even\]). In any other situation, cardinality of $I_{2r}$ will 