Codeforces 598A Tricky Sum【计数】

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：给定一个数n，把1-n里面2的幂次数变为负数。求n个数之和。    分析：先用公式求出1-n的和再减去两倍的1-n以内的2的幂次数    吐槽：QAQ 无语啊。看错了范围，撸了一发Java大数AC了，回头一看，不会爆long long..*/#include<cstdio>#include<iostream>using namespace std;typedef long long ll;int main(){    int t,n;    cin>>t;    while(t--){        scanf("%d",&n);        ll ans = (ll)(1+n)*n/2;        for(ll c=1;c<=n;c<<=1){            ans -=2*c;        }        printf("%I64d\n",ans);    }    return 0;}