CF 598A Tricky Sum【规律】

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.

求一下1到n的和  再减去2的倍数。即为所得

#include <iostream>  #include <cstdio>  #include <cstring>  #include <cmath>  #include <cstdlib>   #include <queue>  #include <stack> #include <algorithm>#define Wi(a) while((a)--) #define Si(a) scanf("%d", &a)  #define Si64(a) scanf("%I64d", &a)#define Pi(a) printf("%d\n", (a)) #define Pl(a) printf("%I64d\n", (a)) #define CLR(a, b) memset(a, (b), sizeof(a)) #define INF 0x3f3f3f3f   #define LL long long using namespace std;int main(){int t;   Si(t);Wi(t){__int64 n, ans;  Si64(n);ans = n*(n+1)/2;for(int i = 0; pow(2, i) <= n; ++i){ans -= (2*(pow(2, i)));}Pl(ans);}return 0;}