Create a sequence of numbers using the "Blum Blum Shub" pseudo-random number generator:

$$\begin{array}{rl}{s}_0& =14025256\\ s_{n+1}& =s_n^{2}mod20300713\end{array}$$

Concatenate these numbers $s_0{s}_1{s}_2\cdots$ to create a string $w$ of infinite length.
Then, $w=14025256741014958470038053646\cdots$

For a positive integer $k$, if no substring of $w$ exists with a sum of digits equal to $k$, $p(k)$ is defined to be zero. If at least one substring of $w$ exists with a sum of digits equal to $k$, we define $p(k)=z$, where $z$ is the starting position of the earliest such substring.

For instance:

The substrings $1,14,1402,\dots$
with respective sums of digits equal to $1,5,7,\dots$
start at position $\mathbf{1}$, hence $p(1)=p(5)=p(7)=\cdots =\mathbf{1}$.

The substrings $4,402,4025,\dots$
with respective sums of digits equal to $4,6,11,\dots$
start at position $\mathbf{2}$, hence $p(4)=p(6)=p(11)=\cdots =\mathbf{2}$.

The substrings $02,0252,\dots$
with respective sums of digits equal to $2,9,\dots$
start at position $\mathbf{3}$, hence $p(2)=p(9)=\cdots =\mathbf{3}$.

Note that substring $025$ starting at position $\mathbf{3}$, has a sum of digits equal to $7$, but there was an earlier substring (starting at position $\mathbf{1}$) with a sum of digits equal to $7$, so $p(7)=1$, not $3$.

We can verify that, for $0<k\le {10}^3$, $\sum p(k)=4742$.

Find $\sum p(k)$, for $0<k\le 2\times {10}^{15}$.