CF1717E.Madoka and The Best University

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!

魔笛想要进入“新西伯利亚国立大学”，但在入学考试中她遇到了一道非常困难的题目：

给定一个整数 $n$，要求计算对所有满足 $a + b + c = n$ 的正整数三元组 $(a, b, c)$，表达式 $\sum{\operatorname{lcm}(c, \gcd(a, b))}$ 的值。

本题中，$\gcd(x, y)$ 表示 $x$ 与 $y$ 的最大公约数，$\operatorname{lcm}(x, y)$ 表示 $x$ 与 $y$ 的最小公倍数。

请帮魔笛解决这道题，助她进入这所顶尖大学！

The first and the only line contains a single integer $n$ ($3 \le n \le 10^5$).

第一行且唯一一行包含一个整数 $n$（$3 \le n \le 10^5$）。

Print exactly one interger — $\sum{\operatorname{lcm}(c, \gcd(a, b))}$. Since the answer can be very large, then output it modulo $10^9 + 7$.

输出一个整数——$\sum{\operatorname{lcm}(c, \gcd(a, b))}$。由于答案可能非常大，因此需对 $10^9 + 7$ 取模后输出。

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, $\operatorname{lcm}(1, \gcd(3, 1)) + \operatorname{lcm}(1, \gcd(2, 2)) + \operatorname{lcm}(1, \gcd(1, 3)) + \operatorname{lcm}(2, \gcd(2, 1)) + \operatorname{lcm}(2, \gcd(1, 2)) + \operatorname{lcm}(3, \gcd(1, 1)) = \operatorname{lcm}(1, 1) + \operatorname{lcm}(1, 2) + \operatorname{lcm}(1, 1) + \operatorname{lcm}(2, 1) + \operatorname{lcm}(2, 1) + \operatorname{lcm}(3, 1) = 1 + 2 + 1 + 2 + 2 + 3 = 11$

在第一个例子中，只有一个合适的三元组 $(1, 1, 1)$。因此答案为 $\operatorname{lcm}(1, \gcd(1, 1)) = \operatorname{lcm}(1, 1) = 1$。

在第二个例子中，
$\operatorname{lcm}(1, \gcd(3, 1)) + \operatorname{lcm}(1, \gcd(2, 2)) + \operatorname{lcm}(1, \gcd(1, 3)) + \operatorname{lcm}(2, \gcd(2, 1)) + \operatorname{lcm}(2, \gcd(1, 2)) + \operatorname{lcm}(3, \gcd(1, 1)) = \operatorname{lcm}(1, 1) + \operatorname{lcm}(1, 2) + \operatorname{lcm}(1, 1) + \operatorname{lcm}(2, 1) + \operatorname{lcm}(2, 1) + \operatorname{lcm}(3, 1) = 1 + 2 + 1 + 2 + 2 + 3 = 11$