CodeForces

传送门：

http://codeforces.com/problemset/problem/598/A

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

Copy

2
4
1000000000

Output

Copy

-4
499999998352516354

Note

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

分析：

题意：

先输入一个T，表示有几组输入，再输入一个n，表示要计算n的和sum。

计算规则为，从1---n这n个数中任意一个数m，如果m这个数是2的次方的话，sum+=(-m)，否则sum+=m。

思路：

先求出总的和sum，再用快速幂，求出所有是2的次方的数的和num，最后求差即可sum=sum-2*num。

code：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
LL qm(int n,int m)//快速幂
{
    LL s=1,x=n;
    while(m)
    {
        if(m&1)
        {
            s*=x;
        }
        x*=x;
        m>>=1;
    }
    return s;
}
int main()
{
    int t;
    cin>>t;
    LL n;
    LL sum;
    while(t--)
    {
        scanf("%I64d",&n);
        sum=n*(n+1)/2;//等差数列求和公式
        for(int i=0;qm(2,i)<=n;i++)
        {
            sum-=(2*qm(2,i));
        }
        printf("%I64d
",sum);
    }
    return 0;
}