## Decreasing Srrnmieeda October Cook off challenge Solution | Easy Explain Decreasing Srrmieeda-

Easy Explanation of Decreasing Srrnmieeda October Cook-off challenge Solution  With Code.Here Is Problem Statement.

## Problem Statement-

You are given two integers $L$ and $R$. Find the smallest non-negative integer $N$ such that

$N\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}\phantom{\rule{thinmathspace}{0ex}}L>N\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}\phantom{\rule{thinmathspace}{0ex}}\left(L+1\right)>\dots >N\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}\phantom{\rule{thinmathspace}{0ex}}\left(R-1\right)>N\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}\phantom{\rule{thinmathspace}{0ex}}R\phantom{\rule{thinmathspace}{0ex}}.$

Here, $\mathrm{%}$ is the modulo operator, so $A\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}\phantom{\rule{thinmathspace}{0ex}}B$ is the remainder of $A$ after division by $B$. For example, $11\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}\phantom{\rule{thinmathspace}{0ex}}3=2$.

### Input

• The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
• The first and only line of each test case contains two space-separated integers $L$ and $R$.

### Output

For each test case, print a single line containing one integer ― the smallest possible $N$, or $-1$ if no solution exists.

It is guaranteed that when a solution exists, the smallest solution does not exceed ${10}^{18}$.

### Constraints

• $1\le T\le {10}^{5}$
• $1\le L

### Example Input

2
4 6
1 2


### Example Output

6
-1


### Explanation

Example case 1: $N=6$ satisfies the given condition, since $6\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}\phantom{\rule{thinmathspace}{0ex}}4\phantom{\rule{thinmathspace}{0ex}}\left(=2\right)>6\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}\phantom{\rule{thinmathspace}{0ex}}5\phantom{\rule{thinmathspace}{0ex}}\left(=1\right)>6\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}\phantom{\rule{thinmathspace}{0ex}}6\phantom{\rule{thinmathspace}{0ex}}\left(=0\right)$. Notice that $N=7$ also satisfies the condition, but it is larger.

Example case 2: It is impossible to find a valid solution because for any non-negative integer $N$, we have $N\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}\left(=0\right)\le N\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}\phantom{\rule{thinmathspace}{0ex}}2$.

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### Decreasing Srrnmieeda  Solution -

In this Problem  given that N%L>N%(L+1)>>N%(R1)>N%R.  and L< R

1-So for satisfying first relationship N should be between >=R otherwise condition one cannot be satisfied.
2-But we know that if N is divisible by L so we  if N> R it is not possible to obtain the relation 1 that is
N%L>N%(L+1)>>N%(R1)>N%R.

According to those two statements, What can be the max value of N%L? Maxmimum value of N%L cannot
exceed 2*L so N<2*L.
and if if(R >= 2*L) we cannot obtain sequence .

so Final Logic is -
if(R >= 2*L) cout << -1 << '\n'; else cout << R << '\n';