Starting from zero the natural numbers are written down in base $10$ like this:
$$0123456789101112\cdots$$

Consider the digit $d=1$. After we write down each number $n$, we will update the number of ones that have occurred and call this number $f(n,1)$. The first values for $f(n,1)$, then, are as follows:

$$\begin{array}{cc}n& f(n,1)\\ 0& 0\\ 1& 1\\ 2& 1\\ 3& 1\\ 4& 1\\ 5& 1\\ 6& 1\\ 7& 1\\ 8& 1\\ 9& 1\\ 10& 2\\ 11& 4\\ 12& 5\end{array}$$

Note that $f(n,1)$ never equals $3$.
So the first two solutions of the equation $f(n,1)=n$ are $n=0$ and $n=1$. The next solution is $n=199981$.

In the same manner the function $f(n,d)$ gives the total number of digits $d$ that have been written down after the number $n$ has been written.
In fact, for every digit $d\ne 0$, $0$ is the first solution of the equation $f(n,d)=n$.

Let $s(d)$ be the sum of all the solutions for which $f(n,d)=n$.
You are given that $s(1)=22786974071$.

Find $\sum s(d)$ for $1\le d\le 9$.

Note: if, for some $n$, $f(n,d)=n$ for more than one value of $d$ this value of $n$ is counted again for every value of $d$ for which $f(n,d)=n$.