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
2017-07-01 21:49:44
take care of perfect squares
2017-06-30 15:41:27
for those getting WA
some testcases:
1 -> 0
99999 -> 48513
500000 -> 730453
9689 -> 1
12345 -> 7431
2017-06-23 08:18:24
it is showing TLE in C please help
2017-06-20 01:07:09
Answer fits in long long for C++.
2017-06-19 01:31:22
what are the test cases i am getting wrong answer?
2017-05-31 05:29:48
Guys I am getting TLE in python I don't know why please help
2017-05-29 19:49:34
Guys take care of n = 1, the solution is 0, cost me 1 WA!
2017-04-29 00:28:23
<1. Don't post any source code here. >
i dont know why judge is telling my answer wrong anybody help


Last edit: 2017-04-29 05:57:29
2017-04-19 22:04:01
for num=1, ans =0, for perfect square count divisor once only
2017-04-03 13:37:13
Simple Problem..!! take care for n=1 ans should be 0 Cost me 2 WA's..!!
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