题目大意：给出t个区间l,r，要求在每个区间内找到两个数a,b满足l<=a+b<=r，且gcd(a,b)!=1
1<=t<=500；1<=l<=r<=1e7
思路：记个结论：gcd(a,b-a)=gcd(a,b)
然后如果区间内有偶数就将其平分，如果只有奇数，就取一个非1的因数即可满足条件，因数和他倍数的gcd就是那个因数
//#include<__msvc_all_public_headers.hpp>
#include<bits/stdc++.h>
using namespace std;
const int N = 500 + 5;
typedef long long ll;
pair<int, int>a[105];
int n;
void init()
{
}
void solve()
{
int l, r;
cin >> l >> r;
if (l!=r||(l==r&&r%2==0))
{
if (r <4)
{//偶数不能是2，因为不能拆出1
cout << -1 << endl;
return;
}
if (r & 1)
{
cout << (r - 1) / 2 << " " << (r - 1) / 2 << endl;
}
else
{
cout << r / 2 << " " << r / 2 << endl;
}
return;
}
else
{
for (int i = 3; i * i <= l; i++)
{
if (l % i == 0)
{
cout << i << " " << l - i << endl;
return;
}
}
cout << -1 << endl;
}
init();
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--)
{
solve();
}
}