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


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


  • 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.


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 1018.


  • 1T105
  • 1L<R106

Example Input

4 6
1 2

Example Output



Example case 1: N=6 satisfies the given condition, since 6%4(=2)>6%5(=1)>6%6(=0). 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%1(=0)N%2.

Problem Link-

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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 

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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