Traveling around the world you noticed that many shop owners raise prices to inadequate values if the see you are a foreigner.

You define inadequate numbers as follows:

• all integers from $1$ to $9$ are inadequate;
• for an integer $x \ge 10$ to be inadequate, it is required that the integer $\lfloor x / 10 \rfloor$ is inadequate, but that's not the only condition. Let's sort all the inadequate integers. Let $\lfloor x / 10 \rfloor$ have number $k$ in this order. Then, the integer $x$ is inadequate only if the last digit of $x$ is strictly less than the reminder of division of $k$ by $11$.

Here $\lfloor x / 10 \rfloor$ denotes $x/10$ rounded down.

Thus, if $x$ is the $m$-th in increasing order inadequate number, and $m$ gives the remainder $c$ when divided by $11$, then integers $10 \cdot x + 0, 10 \cdot x + 1 \ldots, 10 \cdot x + (c - 1)$ are inadequate, while integers $10 \cdot x + c, 10 \cdot x + (c + 1), \ldots, 10 \cdot x + 9$ are not inadequate.

The first several inadequate integers are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 21, 30, 31, 32 \ldots$. After that, since $4$ is the fourth inadequate integer, $40, 41, 42, 43$ are inadequate, while $44, 45, 46, \ldots, 49$ are not inadequate; since $10$ is the $10$-th inadequate number, integers $100, 101, 102, \ldots, 109$ are all inadequate. And since $20$ is the $11$-th inadequate number, none of $200, 201, 202, \ldots, 209$ is inadequate.

You wrote down all the prices you have seen in a trip. Unfortunately, all integers got concatenated in one large digit string $s$ and you lost the bounds between the neighboring integers. You are now interested in the number of substrings of the resulting string that form an inadequate number. If a substring appears more than once at different positions, all its appearances are counted separately.