Considering $4$-digit primes containing repeated digits it is clear that they cannot all be the same: $1111$ is divisible by $11$, $2222$ is divisible by $22$, and so on. But there are nine $4$-digit primes containing three ones: $$1117,1151,1171,1181,1511,1811,2111,4111,8111.$$

We shall say that $M(n,d)$ represents the maximum number of repeated digits for an $n$-digit prime where $d$ is the repeated digit, $N(n,d)$ represents the number of such primes, and $S(n,d)$ represents the sum of these primes.

So $M(4,1)=3$ is the maximum number of repeated digits for a $4$-digit prime where one is the repeated digit, there are $N(4,1)=9$ such primes, and the sum of these primes is $S(4,1)=22275$. It turns out that for $d=0$, it is only possible to have $M(4,0)=2$ repeated digits, but there are $N(4,0)=13$ such cases.

In the same way we obtain the following results for $4$-digit primes.

Digit, d	M(4, d)	N(4, d)	S(4, d)
0	2	13	67061
1	3	9	22275
2	3	1	2221
3	3	12	46214
4	3	2	8888
5	3	1	5557
6	3	1	6661
7	3	9	57863
8	3	1	8887
9	3	7	48073

For $d=0$ to $9$, the sum of all $S(4,d)$ is $273700$.

Find the sum of all $S(10,d)$.