CodeForces 598 A Tricky Sum【数学】

A. 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

241000000000

output

-4499999998352516354

Note

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

题意：

在1-n中，2的幂都为负数，其他为正，求这n个数的和

题解：

求出前n项中2的所有幂的和，然后用前n项的和减去二倍的这个数，得到的结果即为所求。

#include<cstdio>#include<math.h>typedef long long ll;int main(){    int t;    scanf("%d",&t);    while(t--)    {        ll n,tp=1;        scanf("%I64d",&n);        while(tp<=n)        {            tp*=2;        }        --tp;        printf("%I64d\n",n*(n+1)/2-2*tp);    }    return 0;}