# PROBLEM LINK:

Contest Division 1

Contest Division 2

Contest Division 3

Contest Division 4

Setter: Sandeep V

Tester: Istvan Nagy, Aryan

Editorialist: Vijay

# DIFFICULTY:

Simple

# PREREQUISITES:

None

# PROBLEM:

Dazzler has a task for you.

Given two **positive** integers A and B, find two **positive** integers i and j such that:

- gcd(i,j) \gt 1;
- A \leq i \lt j \leq B;
- The value (i + j) is
**minimum**possible.

If there are multiple solutions, you may print any of them. If no such pair exists, print -1.

# EXPLANATION:

*Observation* : We can show that for intervals of length greater than or equal to 4, we can always find two numbers such that their gcd is greater than 1.

Suppose we have an interval of length greater than or equal to 4 . Now,if the interval starts from x , we can write the first 4 numbers as (x,x+1,x+2,x+3). Now, if x is even, then we have (x, x+2) as even, otherwise we have (x+1, x+3) as even. Thus, we have a valid pair of (A, B) with sum **at most** 2 \cdot x+4. Now, if any of A or B is greater than or equal to x+4, we will have the sum greater than or equal to 2 \cdot x + 4. Thus, this will not further improve the answer. So, we can simply ignore and truncate the interval up to length 4.

Now we can just brute force on this truncated interval which can have at most length 4 to obtain the most optimal pair.

# TIME COMPLEXITY:

O(log(min(a,b)) for each test case.

# SOLUTION:

## Editorialist's Solution

```
#include <bits/stdc++.h>
using namespace std;
#define nline '\n'
void solve()
{
int a,b;
cin>>a>>b;
int sum=INT_MAX;
pair<int,int>ans={-1,-1};
for(int i=a;i<=min(b,a+3);i++){
for(int j=i+1;j<=min(b,a+3);j++){
if(__gcd(i,j)>1){
if(i+j<sum){
sum=i+j;
ans={i,j};
}
}
}
}
if(ans.first==-1)cout<<-1<<endl;
else cout<<ans.first<<" "<<ans.second<<endl;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t=1;
cin >> t;
while (t--)
{
solve();
}
}
```