DIVSUM - Divisor Summation

Given a natural number n (1 <= n <= 500000), please output the summation of all its proper divisors.

Definition: A proper divisor of a natural number is the divisor that is strictly less than the number.

e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.

Input

An integer stating the number of test cases (equal to about 200000), and that many lines follow, each containing one integer between 1 and 500000 inclusive.

Output

One integer each line: the divisor summation of the integer given respectively.

Example

Sample Input:
3
2
10
20

Sample Output:
1
8
22

Warning: large Input/Output data, be careful with certain languages


Added by:Neal Zane
Date:2004-06-10
Time limit:3s
Source limit:5000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All
Resource:Neal Zane

hide comments
2020-04-20 17:42:11
When I submit code in python3 for this problem, I get TLE but same logic when I write in cpp, it accepts. Why is this so?

Last edit: 2020-04-20 18:51:41
2020-01-21 17:26:26
for n=1 output is 0
2019-08-08 12:24:56
corner case: n = 1
2019-08-02 15:17:47
same solution in c++ give wrong answer and in c it is Ac
2019-07-27 13:54:01
did anyone know name of file input output please tell me
2019-07-02 06:11:34
Try DIVSUM2 :)
2019-04-21 19:56:37
if the number is 1 , the output must be 0.
fml spent so much time on this stupid question!!
2019-04-15 08:19:39
it was an easy but remember answer of 1 is 0 always.
2019-04-13 09:56:13
AC in one shot use i*i<=num to avoid time isssue in this problem
2019-03-15 15:08:14
TOO easy, time limit should have been less
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