Sumdiv
时间限制:C/C++/Rust/Pascal 1秒,其他语言2秒
空间限制:C/C++/Rust/Pascal 32 M,其他语言64 M
64bit IO Format: %lld

题目描述

Consider two natural numbers A and B. Let S be the sum of all natural divisors of . Determine S modulo 9901 (the rest of the division of S by 9901).

输入描述:

The only line contains the two natural numbers A and B, (0 <= A,B <= 50000000)separated by blanks.

输出描述:

The only line of the output will contain S modulo 9901.
示例1

输入

复制
2 3

输出

复制
15

说明

2^3 = 8.
The natural divisors of 8 are: 1,2,4,8. Their sum is 15.
15 modulo 9901 is 15 (that should be output).