Sum $sum_{(k_1, k_2, k_3): k_1+k_2+k_3=K, ,, n_1+n_2+n_3=N}k_1^{n_1}times k_2^{n_2} times k_3^{n_3}$

Let $k_1, k_2, k_3$ be natural non-negative numbers such that $k_1+k_2+k_3=K$. Let $n_1, n_2, n_3 in {0, ldots, N}$ and such that $n_1+n_2+n_3=N$.

Calculate

$$
S=sum_{(k_1, k_2, k_3): k_1+k_2+k_3=K, ,, n_1+n_2+n_3=N}k_1^{n_1}times k_2^{n_2} times k_3^{n_3}
$$

My attempt:
I am thinking on representing this sum as a chain of sums over each summand $k_j$. For example, the interior sum would be:
$
sum_{k_3=0}^{K-k_1-k_2}k_3^{n_3}.
$
Using Sums of p-th powers formula we can get $$sum_{k_3=0}^{K-k_1-k_2}k_3^{n_3}=frac{B_{n_3+1}(K-k_1-k_2+1)-B_{n_3+1}}{n_3}.$$ So, the sum $S$ would be represented as a product of these ratios with Bernoulli numbers $B_n$.

Is there a better way on computing/estimating from above sum $S$?