Tricky Sum

Tricky Sum

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.

Example

Input

241000000000

Output

-4499999998352516354

Note

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

等差数列和等比数列的运用，容易想到等差，不容易想到等比

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