C. Sum in Binary Tree

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vanya really likes math. One day when he was solving another math problem, he came up with an interesting tree. This tree is built as follows.

Initially, the tree has only one vertex with the number 11 — the root of the tree. Then, Vanya adds two children to it, assigning them consecutive numbers — 22 and 33, respectively. After that, he will add children to the vertices in increasing order of their numbers, starting from 22, assigning their children the minimum unused indices. As a result, Vanya will have an infinite tree with the root in the vertex 11, where each vertex will have exactly two children, and the vertex numbers will be arranged sequentially by layers.

Part of Vanya's tree.

Vanya wondered what the sum of the vertex numbers on the path from the vertex with number 11 to the vertex with number n� in such a tree is equal to. Since Vanya doesn't like counting, he asked you to help him find this sum.

Input

The first line contains a single integer t� (1≤t≤1041≤�≤104) — the number of test cases.

This is followed by t� lines — the description of the test cases. Each line contains one integer n� (1≤n≤10161≤�≤1016) — the number of vertex for which Vanya wants to count the sum of vertex numbers on the path from the root to that vertex.

Output

For each test case, print one integer — the desired sum.

Example

input

Copy

6

3

10

37

1

10000000000000000

15

output

Copy

4
18
71
1
19999999999999980
26

Note

In the first test case of example on the path from the root to the vertex 33 there are two vertices 11 and 33, their sum equals 44.

In the second test case of example on the path from the root to the vertex with number 1010 there are vertices 11, 22, 55, 1010, sum of their numbers equals 1+2+5+10=181+2+5+10=18.

 

解题说明：此题是一道数学题，找规律能发现父节点为叶子节点除以2，那就从叶节点除非，不断除以2，直到根节点1为止。

#include
int main()
{
	long long int t, i, n, p = 0;
	scanf("%lld", &t);
	while (t--)
	{
		scanf("%lld", &n);
		p = n;
		while (n != 1)
		{
			p = p + n / 2;
			n = n / 2;
		}
		printf("%lld\n", p);
		p = 0;
	}
	return 0;
}