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);}
}

转载于:https://www.cnblogs.com/-xiangyang/p/9220273.html

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