TJUT1 - Divisors

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

Input

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

Output

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

Example

Input:
9 12
1 2147483647
0 0

Output:
15
46475828386

Explanation

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.


Added by:abdou_93
Date:2013-10-16
Time limit:1.182s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64
Resource:TJU

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2023-08-27 23:41:36
any hints ??????
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