题目链接：http://codeforces.com/problemset/problem/598/A
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.

Examples
Input
2
4
1000000000
Output
-4
499999998352516354
Note
The answer for the first sample is explained in the statement.

规律很好找，先把1~n都按正数加起来，用求和公式算一下
再减去2的次幂数的和乘以2
Sn=(1+n)n/2-2(2^(k+1)-1)
2^(k+1)-1是用等比数列求和公式算出来的，这个k是n的二进制最高位上那个1对应是2的k次幂
举例
n k
1 0
2 1

n k
1 0
2 1
3 1
4 2
5 2
– –

<mark>注意pow（）的精度丢失</mark>
代码里的k就是上文的k+1

#include<cstdio>
#include<cmath>
using namespace std;
int main() {
	long long t,n;
	scanf("%lld",&t);
	while(t--){
		scanf("%lld",&n);
		long long sum=(1+n)*n/2;
		int k=0;
		while(n){
			k++;
			n>>=1;
		}
		sum=sum-2*(pow(2,k)-1);
		printf("%lld\n",sum);
	}
	return 0;
}