# Odd Long Tally Counter

This problem is the more-evil-version of this problem.

A normal long tally counter has $$7$$ digits and counts $$1$$ by $$1$$, hence there are $$10^{7}$$ different numbers can be shown on it (in range $$0000000 ... 9999999$$).

Suppose you have an odd long tally counter. Instead $$1$$ by $$1$$, it counts $$x$$ by $$x$$. In case of overflow, the counter only shows the last $$7$$ digits of the number.

For example, if $$x = 1000001$$ the counter will show these numbers : $$0000000, 1000001, 2000002, ..., 9000009, 0000010, 1000011, ...$$

Let $$F(n)$$ be the function returning the number of different numbers can be shown on odd long tally counter with $$x = n$$, find the last $$3$$ digits of $$\sum_{i=1}^{10^7-1} F(i)$$.

As an explicit example : $$F(1) = 10^7$$

This problem is taken from TOKI Open Contest January 2013 (Problemsetter : Me)

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