Madoka and The Best University

题意翻译

对于任意正整数三元组 $(a,b,c)$ 满足 $a+b+c = n$。求 $\sum{\operatorname{lcm}(c, \gcd(a, b))}$ 的值。答案对 $10^9+7$ 取模。

题目描述

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](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $ x $ and $ y $ , and $ \operatorname{lcm}(x, y) $ denotes the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of $ x $ and $ y $ . Solve this problem for Madoka and help her to enter to the best university!

输入输出格式

输入格式

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

输出格式

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 $ .

输入输出样例

输入样例 #1

3

输出样例 #1

1

输入样例 #2

5

输出样例 #2

11

输入样例 #3

69228

输出样例 #3

778304278

说明

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 $