We define the function f(x) = the number of divisors of x. Given two integers a and b (a ≤ b), please calculate f(a) + f(a+1) + ... + f(b).

Input

Two integers a and b for each test case, 1 ≤ a ≤ b ≤ 2³¹ - 1. The input is terminated by a line with a = b = 0.

Output

The value of f(a) + f(a+1) + ... + f(b).

Sample Input

9 12
1 2147483647
0 0

Sample Output

15
46475828386

Hint

For the first test case:
9 has 3 divisors: 1, 3, 9.
10 has 4 divisors: 1, 2, 5, 10.
11 has 2 divisors: 1, 11.
12 has 6 divisors: 1, 2, 3, 4, 6, 12.
So the answer is 3 + 4 + 2 + 6 = 15.

if you find Source code limit is small here you can solve the tutorial version here :
http://www.spoj.com/problems/TJUT1/