Preparando MOJI
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 $$$10^9 + 7$$$.
3
1
5
11
69228
778304278
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, $$$\operatorname{lcm}(1, \gcd(3, 1)) + \operatorname{lcm}(1, \gcd(2, 2)) + \operatorname{lcm}(1, \gcd(1, 3)) + \operatorname{lcm}(2, \gcd(2, 1)) + \operatorname{lcm}(2, \gcd(1, 2)) + \operatorname{lcm}(3, \gcd(1, 1)) = \operatorname{lcm}(1, 1) + \operatorname{lcm}(1, 2) + \operatorname{lcm}(1, 1) + \operatorname{lcm}(2, 1) + \operatorname{lcm}(2, 1) + \operatorname{lcm}(3, 1) = 1 + 2 + 1 + 2 + 2 + 3 = 11$$$
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Provedor Codeforces
Código CF1717E
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Datas 09/05/2023 10:30:52
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