#P1073E. Segment Sum

    ID: 4483 Type: RemoteJudge 1000ms 256MiB Tried: 0 Accepted: 0 Difficulty: (None) Uploaded By: Tags>bitmaskscombinatoricsdpmath*2300

Segment Sum

No submission language available for this problem.

Description

You are given two integers $l$ and $r$ ($l \le r$). Your task is to calculate the sum of numbers from $l$ to $r$ (including $l$ and $r$) such that each number contains at most $k$ different digits, and print this sum modulo $998244353$.

For example, if $k = 1$ then you have to calculate all numbers from $l$ to $r$ such that each number is formed using only one digit. For $l = 10, r = 50$ the answer is $11 + 22 + 33 + 44 = 110$.

The only line of the input contains three integers $l$, $r$ and $k$ ($1 \le l \le r < 10^{18}, 1 \le k \le 10$) — the borders of the segment and the maximum number of different digits.

Print one integer — the sum of numbers from $l$ to $r$ such that each number contains at most $k$ different digits, modulo $998244353$.

Input

The only line of the input contains three integers $l$, $r$ and $k$ ($1 \le l \le r < 10^{18}, 1 \le k \le 10$) — the borders of the segment and the maximum number of different digits.

Output

Print one integer — the sum of numbers from $l$ to $r$ such that each number contains at most $k$ different digits, modulo $998244353$.

Samples

10 50 2

1230

1 2345 10

2750685

101 154 2

2189

Note

For the first example the answer is just the sum of numbers from $l$ to $r$ which equals to $\frac{50 \cdot 51}{2} - \frac{9 \cdot 10}{2} = 1230$. This example also explained in the problem statement but for $k = 1$.

For the second example the answer is just the sum of numbers from $l$ to $r$ which equals to $\frac{2345 \cdot 2346}{2} = 2750685$.

For the third example the answer is $101 + 110 + 111 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 119 + 121 + 122 + 131 + 133 + 141 + 144 + 151 = 2189$.