If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below N.

Input Format
First line contains T that denotes the number of test cases. This is followed by T lines, each containing an integer, N

Output Format
For each test case, print an integer that denotes the sum of all the multiples of 3 or 5 belowN.

Constraints
1≤T≤105
1≤N≤109

Sample Input

2
10
100

Sample Output

23
2318 

Language: C

#include <stdio.h>
 

#include <stdlib.h>

​

int main()

{

    unsigned long long int N,i,j,x,y,z;

    int T;

    scanf("%d",&T);

    unsigned long long int n[T];

    for(i=1;i<=T;i++)

    {

        scanf("%llu",&N);

        x=(N-1)/3;

        y=(N-1)/5;

        z=y/3;

        n[i-1]=(3+3*x)*x/2;

        n
 
[i-1]+=(5+5*y)*y/2;

        n[i-1]-=(15+15*z)*z/2;

    }

    for(i=1;i<=T;i++)

    {

        printf("%llu\n",n[i-1]);

    }

    return 0;

}