Coderforce 598A Tricky Sum (数学)

题意：求数字之和，但是要减去2的倍数

思路：直接求...

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

题目

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

241000000000

Output

-4499999998352516354

Hint

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