The following is a process for finding the number of functions $f : X \rightarrow X$ where $X = \{ 1,2,3,4,5,6 \}$ such that the range of the composite function $f \circ f$ has 5 elements.\\
Let the ranges of functions $f$ and $f \circ f$ be $A$ and $B$, respectively.\\
If $n ( A ) = 6$, then $f$ is a bijection, and $f \circ f$ is also a bijection, so $n ( B ) = 6$.\\
Also, if $n ( A ) \leq 4$, then $B \subset A$, so $n ( B ) \leq 4$.\\
Therefore, we only need to consider the case where $n ( A ) = 5$, that is, $B = A$.\\
(i) The number of ways to choose a subset $A$ of $X$ with $n ( A ) = 5$ is (가).\\
(ii) For the set $A$ chosen in (i), let $k$ be the element of $X$ that does not belong to $A$. Since $n ( A ) = 5$, the number of ways to choose $f ( k )$ from set $A$ is (나).\\
(iii) For $A = \left\{ a _ { 1 } , a _ { 2 } , a _ { 3 } , a _ { 4 } , a _ { 5 } \right\}$ chosen in (i) and $f ( k )$ chosen in (ii), since $f ( k ) \in A$ and $A = B$, we have $A = \left\{ f \left( a _ { 1 } \right) , f \left( a _ { 2 } \right) , f \left( a _ { 3 } \right) , f \left( a _ { 4 } \right) , f \left( a _ { 5 } \right) \right\} \cdots ( * )$. The number of cases satisfying (*) is equal to the number of bijections from set $A$ to set $A$, so it is (다).\\
Therefore, by (i), (ii), and (iii), the number of functions $f$ we seek is (가) $\times$ (나) $\times$ (다).\\
When the numbers corresponding to (가), (나), and (다) are $p$, $q$, and $r$, respectively, what is the value of $p + q + r$? [4 points]\\
(1) 131\\
(2) 136\\
(3) 141\\
(4) 146\\
(5) 151