#P1717E. Madoka and The Best University
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ID: 7891
Type: RemoteJudge
1000ms
256MiB
Tried: 0
Accepted: 0
Difficulty: (None)
Uploaded By:
Hydro
Tags> Madoka and The Best University
No submission language available for this problem.
Description
Madoka wants to enter to "Novosibirsk State University", but in the entrance exam she came across a very difficult task:
Given an integer $n$, it is required to calculate $\sum{\operatorname{lcm}(c, \gcd(a, b))}$, for all triples of positive integers $(a, b, c)$, where $a + b + c = n$.
In this problem $\gcd(x, y)$ denotes the greatest common divisor of $x$ and $y$, and $\operatorname{lcm}(x, y)$ denotes the least common multiple of $x$ and $y$.
Solve this problem for Madoka and help her to enter to the best university!
The first and the only line contains a single integer $n$ ($3 \le n \le 10^5$).
Print exactly one interger — $\sum{\operatorname{lcm}(c, \gcd(a, b))}$. Since the answer can be very large, then output it modulo $10^9 + 7$.
Input
The first and the only line contains a single integer $n$ ($3 \le n \le 10^5$).
Output
Print exactly one interger — $\sum{\operatorname{lcm}(c, \gcd(a, b))}$. Since the answer can be very large, then output it modulo $10^9 + 7$.
3
5
69228
1
11
778304278
Note
In the first example, there is only one suitable triple $(1, 1, 1)$. So the answer is $\operatorname{lcm}(1, \gcd(1, 1)) = \operatorname{lcm}(1, 1) = 1$.
In the second example, $\operatorname{lcm}(1, \gcd(3, 1)) + \operatorname{lcm}(1, \gcd(2, 2)) + \operatorname{lcm}(1, \gcd(1, 3)) + \operatorname{lcm}(2, \gcd(2, 1)) + \operatorname{lcm}(2, \gcd(1, 2)) + \operatorname{lcm}(3, \gcd(1, 1)) = \operatorname{lcm}(1, 1) + \operatorname{lcm}(1, 2) + \operatorname{lcm}(1, 1) + \operatorname{lcm}(2, 1) + \operatorname{lcm}(2, 1) + \operatorname{lcm}(3, 1) = 1 + 2 + 1 + 2 + 2 + 3 = 11$