Given a and b up to 1e12 with b-a at most 1e6, sum the divisor sums F(n) over every n in [a, b].
The Slivians are afraid of factoring. It is, well, difficult.
They do not even care about the factors themselves, only about how much they sum to.
Define F(n)F(n)F(n) as the sum of all factors of nnn, so F(6)=12F(6) = 12F(6)=12 and F(12)=28F(12) = 28F(12)=28. Given two integers aaa and bbb with a≤ba \le ba≤b, compute
S=∑a≤n≤bF(n)S=\sum_{a \le n \le b} F(n)S=∑a≤n≤bF(n)
The first line contains two integers aaa and bbb separated by a space. (1≤a≤b≤10121 \le a \le b \le 10^{12}1≤a≤b≤1012, b−a≤106b - a \le 10^{6}b−a≤106)
Print SSS on a single line.