Chef's new friend hErd gave him two functions f and g.

The function f is defined over x (x≥ 1) as:

$f(x) = \begin{cases}$
0, & $\text{if$ } x = 1 \
$f( \frac{x}{2} ) + 1$, & $\text{if$ $} x \text{$ is even} \
$f( \lfloor \frac{x}{2} \rfloor )$, & $\text{if$ $} x \text{$ is odd} \
$\end{cases}$

The function g is defined over x (x≥ 1) as:

$g(x) = \begin{cases}$
1, & $\text{if$ } x = 1 \
$2\cdot g( \frac{x}{2} ) + 1$, & $\text{if$ $} x \text{$ is even} \
$2\cdot g( \lfloor \frac{x}{2} \rfloor )$, & $\text{if$ $} x \text{$ is odd} \
$\end{cases}$

where $\lfloor z \rfloor,$ denotes the greatest integer less than or equal to z.

He also gave Chef two integers L and R. Chef has to find the maximum value of f(x)+g(x) for L ≤ x ≤ R.

------ Input Format ------

• The first line of input will contain a single integer T, denoting the number of test cases.
• The only line of each test case contains two space-separated integers L and R, as mentioned in the statement.

------ Output Format ------

For each test case, output on a new line the maximum value of f(x)+g(x) for L ≤ x ≤ R.

------ Constraints ------

$1 ≤ T ≤ 10^{5}$
$1 ≤ L ≤ R ≤ 10^{9}$

----- Sample Input 1 ------
3
1 1
1 2
1 20
----- Sample Output 1 ------
1
4
35
----- explanation 1 ------
Test case $1$: $f(1)=0$ and $g(1)=1$. Hence, $f(x) + g(x) = 1$.

Test case $2$: There are $2$ possible values of $x$.
- $x = 1$: $f(1)+g(1)=1$
- $x = 2$: $f(2)+g(2)=(f(1) + 1) + (2\cdot g(1) + 1) = 1 + 3 = 4$.

Hence the maximum value of $f(x) + g(x) = 4$.

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