CodeForces 598A Tricky Sum

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

2
4
1000000000

Output

-4
499999998352516354

Hint

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

数学题总是有技巧，不能总是暴力解。

#include<cstdio>
#define ll long long
int main()
{
ll t;
scanf("%lld",&t);
while(t--)
{
ll n,i;
scanf("%lld",&n);
ll sum = (1 + n)*n/2; //先求出n个数的和
for(i = 1 ; i <= n ; i*=2)
{
sum -= (i*2); //减去2的k次方的2倍
// printf("*****%lld\n",sum);
}
printf("%lld\n",sum);
}
}