Coderforce 598A Tricky Sum (数学)

题意：求数字之和，但是要减去2的倍数

思路：直接求...

#include<cstdio>
#include <cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

int main ()
{
long long sum,n,t,x;
int i;
scanf("%lld",&t);
while(t--)
{
scanf("%lld",&n);
sum=n*(n+1)/2;
x=sum;
for(i=0;pow(2,i)<=n;i++)
{
sum-=(2*pow(2,i));
}
printf("%lld\n",sum);
}
return 0;
}

题目

Description

In this problem you are to calculate the sum of all integers from 1 to n, but you should take all powers of two with minus in the sum.

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

Calculate the answer for t values of n.

Input

The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of values of n to
be processed.

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

Output

Print the requested sum for each of t integers n given in the input.

Sample Input

Input

2
4
1000000000

Output

-4
499999998352516354

Hint

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