求和减2的次方

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.

代码：

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<string.h>
using namespace std;
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
while(n--)
{
long long a,i,ans=0;
scanf("%lld",&a);
for(i=1;i<=a;i=i*2)
{
ans-=2*i;
}
ans+=(a*(a+1)/2);
printf("%lld\n",ans);
}
}
return 0;
}

题意：给你一个数n，计算从一到n的整数和。不过如果其中的数是2的x次方时，不仅不加而且要减去。输出总和。

思路：一个for循环判断是不是2的次方，不用i++，直接i=2*i.其余的数直接用求和公式