cf 598 A

A. Tricky Sum

time limit per test

memory limit per test

input

output

1 to n, but you should take all powers of two with minus in the sum.

n = 4 the sum is equal to  - 1 - 2 + 3 - 4 =  - 4, because 1, 2 and 4 are 20, 21 and 22

t values of n.

Input

t (1 ≤ t ≤ 100) — the number of values of n

t lines contains a single integer n (1 ≤ n ≤ 109).

Output

t integers n

Sample test(s)

input

2 4 1000000000

output

-4 499999998352516354

Note

The answer for the first sample is explained in the statement.

#include <bits/stdc++.h>
using namespace std;


int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        __int64 n,sum,res=1,a=1;
        scanf("%I64d",&n);
        sum=(n*(n+1))/2;
        for(__int64 i=1;i<35;i++)
        {
            a*=2;
            if(a<=n)
                res+=a;
            else
                break;
        }
        printf("%I64d\n",sum-2*res);
    }
    return 0;
}