【CodeForces - 598A】Tricky Sum

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                                                                                                           Tricky
Sum

In this problem you are to calculate the sum of all integers from 1to 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, because1, 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.

Example

Input

2
4
1000000000

Output

-4
499999998352516354

Note

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

//            有些编译器不接受long long 所以用_int64 或头文件改成C++

//代码如下：

#include <cstdio>
int main()
{
long long int sum,mul,ans,k;
long long int n;
int t;
scanf ("%d",&t);
while(t--)
{
ans = 1;
mul = 1;
k = 1;
scanf ("%lld",&n);
sum = n * (n+1) / 2;
while (k <= n)
{
k*=2;
ans++;
}
// for (long long i = 1 ; i <= ans ; i++)
// {
// mul *= 2;
// }
k = k - 1;
sum = sum - 2*k;
printf ("%lld\n",sum);
}
return 0;
}