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=;
     __int64 a=;
     int ans=;
     cin>>t;
     while(t--)
     {
         ans=;
         sum=;
         a=;
         cin>>n;
         sum+=(+n)*n/;
         //cout<<sum<<endl;
         while(ans<=n)
         {
             sum-=*ans;
             ans=*ans;
         }
        // cout<<a<<endl;
         cout<<sum<<endl;
     }
     return ;
 }