A problem is easy

A problem is easy

Description

When Teddy was a child, he was always thinking about some simple math problems, such as "what it's 1 cup of water plus 1 pile of dough .., "100 yuan buy 100 pig ". etc ..

One day Teddy met a old man in his dream, in that dream the man whose name was "rulai" gave teddy a problem:
Given an N, Can you calculate how many ways to write N as I * j + I + J (0 <I <= J )?
Teddy found the answer when N was less than 10... But if n get bigger, he found it was too difficult for him to solve. Well, you clever acmers, cocould you help little Teddy to solve this problem and let him have a good dream?

 

Input

The first line contain a T (t <= 2000). Followed by T lines, each line contain an integer N (0 <= n <= 10 ^ 11 ).

Output

For each case, output the number of ways in one line

Sample Input

213

Sample output

01

 #include <iostream>#include <cmath>using namespace std;int main(){    int n;    int a,b;    cin>>n;    while(n--)    {        cin>>a;        b=0;        for(int i=2;i*i<=(a+1);i++)        {            int j = (a+1)/i;            if(i*j==a+1 && i<=j)                        b++;        }        cout<<b<<endl;    }return 0;}        

 

A problem is easy