&= \frac{a_1}{2^{\frac{p+q-4}{2}}  } (\E X^4_1- 1),\end{aligned}$$ where $a_1=(\binom{p}{p-2}\binom{p-2}{\frac{p-2}{2}} (\frac{p-2}{2})!)(\binom{q}{q-2}\binom{q-2}{\frac{q-2}{2}} (\frac{q-2}{2})!)$. Therefore from (\[eqn:lim\_cov\_SC\]), (\[eqn:p,q\_even1\]) and (\[eqn:p,q\_even2\]), we get $$\begin{aligned}
 \label{eqn:p,q_even}
  \lim_{n\to\infty} &\frac{1}{n^{\frac{p+q}{2}-1}} \sum_{r=1}^{ \min\{ \frac{p}{2},\frac{q}{2} \} } \sum_{I_{2r}} \big( \E[X_{i_1} \cdots X_{i_{p}} X_{j_1} \cdots X_{j_{q}} ] -\E[ X_{i_1} \cdots X_{i_{p}} ] \E[ X_{j_1} \cdots X_{j_{q}} ]  \big)  \\
  &= \frac{a_1}{2^{\frac{p+q-4}{2}}  } (\E X^4_1- 1) + \sum_{r=2}^{ \min\{ \frac{p}{2},\frac{q}{2} \} } \frac{a_r}{2^{\frac{p+q-4r}{2}}  } \sum_{s=0}^{2r}\binom{2r}{s}^2 s!(2r-s)! \ h_{2r}(s).\nonumber\end{aligned}$$

**subcase II.** **$p,q$ both are odd:** In this case we calculate right hand side of (\[eqn:lim\_cov\_SC\]) for odd value of $p, q$ and $m$. If $m=2r+1$ for $r=0,1, \ldots, \min\{\frac{p-1}{2}, \frac{q-1}{2}\}$, then the typical term of $I_{2r+1}$ looks like $$((d_1, d_2,\ldots, d_{2r+1}, i_{2r+2}, \ldots, i_{p}), (d_1,d_2,\ldots, d_{2r+1}, j_{2r+2}, \ldots, j_{q}))$$ and for such an elements of $I_{2r+1}$, the number of free entries in $I_{2r+1}$ will be maximum, if following conditions hold

1.  each entries of $ \{d_1, d_2, \ldots, d_{2r+1}\}$ are distinct,

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

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

and the contribution will be of the order $O(n^{2r+1-1 +\frac{p-2r-1}{2} +\frac{q-2r-1}{2} })= O(n^{\frac{p+q}{2} -1})$, where $(-1)$ is arising due to the constraint, $\sum_{t=1}^{2r+1}\epsilon_t d_t =0\; \mbox{(mod n}).$ In any other 