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*Counting Digits (noch nicht übersetzt)*

Starting from zero the natural numbers are written down in base 10 like this:

0 1 2 3 4 5 6 7 8 9 10 11 12....

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:

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 |

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` ≠ 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 ∑ `s`(`d`) for 1 ≤ d ≤ 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`.