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

 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.$$N\mathrm{\%}L>N\mathrm{\%}(L+1)>\dots >N\mathrm{\%}(R-1)>N\mathrm{\%}R.$$

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

Input

• 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$1\le L<R\le {10}^6$

Example Input

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)$6\mathrm{\%}4(=2)>6\mathrm{\%}5(=1)>6\mathrm{\%}6(=0)$. 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$N\mathrm{\%}1(=0)\le N\mathrm{\%}2$.

Problem Link-https://www.codechef.com/COOK123B/problems/DECREM

Join Telegram For Competetive Programming Guide and Update-https://t.me/competitiveProgrammingDiscussion

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-