Tricky Sum

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                                   Tricky Sum

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.

Example

Input

2
4
1000000000

Output

-4
499999998352516354

Note

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

題解：思路就是先把全部的和算出來，然後再減去2的冪次方的數就行了（2的冪次方單獨算）

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
using namespace std;
struct P
{
	int star;
	int endd;
}num[900000];
bool cmp(P a,P b)
{
	return a.endd<b.endd;
}
int main()
{
	int n,k,p;
	while(scanf("%d",&n)!=EOF)
	{
		k=1;
		for(int i=0;i<n;i++)
		{
			scanf("%d%d",&num[i].star,&num[i].endd);
		}
		sort(num,num+n,cmp);
		p=num[0].endd;
		for(int i=1;i<n;i++)
		{
			if(num[i].star>p)
			{
				k++;
				p=num[i].endd;
			}
		}
		printf("%d\n",k);
	}
	return 0;
}