(CodeForces - 598A)Tricky Sum

(CodeForces - 598A)Tricky Sum

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

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.

题目大意：给出数字n，求和，其中遇到2的指数次应该减到，例如n=4，则sum=-1-2+3-4=-4.

思路：先根据求和公式求出1……n的和sum=n*(n+1)/2，再减去2的指数次的两倍（因为之前多加了一次，而要求是减去，所以总共需要减去2倍）。

#include<cstdio>
using namespace std;

typedef long long LL;

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