Educational Codeforces Round 1 A

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 and22 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

2
4
1000000000

output

-4
499999998352516354

Note

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

题意：2,4,这种是2的次方数，我们是减去，其他是加

解法：先全部加起来，再一个个减去2的倍数就好了

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
using namespace std;
int main()
{
    int t;
    __int64 n;
    __int64 sum=0;
    __int64 a=0;
    int ans=1;
    cin>>t;
    while(t--)
    {
        ans=1;
        sum=0;
        a=0;
        cin>>n;
        sum+=(1+n)*n/2;
        //cout<<sum<<endl;
        while(ans<=n)
        {
            sum-=2*ans;
            ans=2*ans;
        }
       // cout<<a<<endl;
        cout<<sum<<endl;
    }
    return 0;
}