codeforces598A（快速幂）

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.

题目大意：1~n中所有2的次方的数都变为负数（从0次方开始算起），最后计算这n项和。

解题思路：首先，在所有2的幂未变负数时，前n和为等差数列的前n项和，1~n内2的幂可以看成等比数列的前k项和（2^k<n）.

最后的结果就是1~n的前n项和减去2倍等比数列的前k项和。

AC代码：

#include<cstdio>#include<cmath>__int64 xpow(__int64 b){__int64 res=1,a=2;while(b>0){if(b&1==1) res*=a;a=a*a;b>>=1;}return res;}int main(){int t;__int64 n,s1,s2,result,i,k;scanf("%d",&t);while(t--){scanf("%I64d",&n);s1=(n*(n+1))/2;i=0;while(1){k=xpow(i);if(k>n) break;i++;}s2=xpow(i)-1;printf("%I64d\n",s1-2*s2);}return 0;}