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


hide comments
jvjplus: 2017-12-16 13:54:55

remember: sqrt(x)*sqrt(x)!=x vs (int)sqrt(x)*(int)sqrt(x)==x

guilheramos: 2017-12-06 14:42:31

Somebody solve this problem using python?

yashasvi_spoj: 2017-12-03 08:35:43

AC?? Help What is this?
_/\_

kilojewels: 2017-12-01 12:27:13

can anyone help me understand hoow to use stdin.read() in python

gopal_05: 2017-11-06 04:14:17

why are we taking square root of n ?

jayharsh: 2017-08-12 14:33:06

AC in one Go....! simple .......

kyrillos: 2017-07-26 19:53:38

easy one solved in 0.06 s

s_harsha: 2017-07-01 21:49:44

take care of perfect squares

sanjitpd_777: 2017-06-30 15:41:27

for those getting WA
some testcases:
1 -> 0
99999 -> 48513
500000 -> 730453
9689 -> 1
12345 -> 7431

aman_99: 2017-06-23 08:18:24

it is showing TLE in C please help


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