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
2016-05-15 21:32:31
tle. any help ?
2016-05-08 21:12:56
yeah...simple maths..:) :) feeling gud after doing it wid just 1 try(actually got 1 wa coz of that n=1 )..yeah :D
2016-04-24 15:43:14
@hrithik,vijay... calculate divisors up to sqrt(n) . For better runtime you can use modified sieve
2016-04-19 19:59:31
getting tle please how do we do this??
2016-03-24 06:12:37
time limit exceeded how to
2016-02-14 18:27:14
Just need to know that divisors exist in pairs. If i divides n then n/i also divides n. And smaller of the 2 divisors lies below sqrt(n)
2016-02-10 18:10:03
How to wite test case in program?
2016-02-01 08:20:43 Abhishek Kumar Singh
For those who are getting WA, take care of the square no.'s.That is no's such as 16.
2016-01-19 16:25:07 Nallagatla Manikanta
For n=1, answer should be 0. Cost me 1 WA!! Got AC finally!
2016-01-17 19:09:07
Hello everybody, Could anybody explain me why runtime error is occuring in my judge, (I have tested the program on my c++ compiler , the answer is being correcly for every case also 1)
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