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.

题意     计算-1-2+3-4+5+6+7-8........这个公式。

解法     将所有2的次方存起来。

#include<cstdio>
long long a[40];
long long pow(int n)
{
	if(n==0)
	return 1;
	else
	{
		long long k1=1;
		for(int i=0;i<n;i++)
		k1*=2;
		return k1;
	}
 
}
int main()
{
	int t;
	scanf("%d",&t);
	for(int i=0;i<33;i++)
	a[i]=pow(i);
	while(t--)
	{
		long long n;
		scanf("%lld",&n);
		long long sum;
		sum=(1+n)*n/2;
		int i;
 
		for( i=0;i<33;i++)
			if(n<=a[i])
			break;
			if(n==a[i])
			i=i+1;
		long long sum1=0;
 
		for(int j=0;j<i;j++)
			sum1+=a[j];
		sum=sum-sum1*2;
		printf("%lld\n",sum);
 
	}
}