## 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$L$ and R$R$. Find the smallest non-negative integer N$N$ such that

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

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

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

### Output

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

It is guaranteed that when a solution exists, the smallest solution does not exceed 1018${10}^{18}$.

### Constraints

- 1≤T≤105$1\le T\le {10}^{5}$
- 1≤L<R≤106$1\le L<R\le {10}^{6}$

```
2
4 6
1 2
```

### Example Output

```
6
-1
```

### Explanation

Example case 1: N=6$N=6$ satisfies the given condition, since 6%4(=2)>6%5(=1)>6%6(=0)$6\phantom{\rule{thinmathspace}{0ex}}\mathrm{\%}\phantom{\rule{thinmathspace}{0ex}}4\phantom{\rule{thinmathspace}{0ex}}(=2)>6\phantom{\rule{thinmathspace}{0ex}}\mathrm{\%}\phantom{\rule{thinmathspace}{0ex}}5\phantom{\rule{thinmathspace}{0ex}}(=1)>6\phantom{\rule{thinmathspace}{0ex}}\mathrm{\%}\phantom{\rule{thinmathspace}{0ex}}6\phantom{\rule{thinmathspace}{0ex}}(=0)$. Notice that N=7$N=7$ 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$N$, we have N%1(=0)≤N%2$N\phantom{\rule{thinmathspace}{0ex}}\mathrm{\%}\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}(=0)\le N\phantom{\rule{thinmathspace}{0ex}}\mathrm{\%}\phantom{\rule{thinmathspace}{0ex}}2$.

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### Decreasing Srrnmieeda Solution -

In this Problem given that N%L>N%(L+1)>…>N%(R−1)>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%(R−1)>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';

### Here is m code-

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