Here is the link to the problem :

What I have done is that first I found all the prime number from 1 to 10000 using the sieve of Eratosthenes. Then, for each element in the array (the list of n numbers), I found all the numbers in the sieve(whose elements are stored in an array) that will divide this element(the element of the array). Now, I found out the maximum power of that prime factor satisfying this. I did this for all the elements of the sieve which are smaller than this element(the element of the array). I did this for all the elements of the array, thus storing the minimum power of each prime factor required to make the LCM.

After this, I traversed from 1 to m and found out the factor with which this each number will multiply the LCM if it is included. In this process, I found out the element that will have the maximum multiplication factor. Now I am unable to think of a proper big O for my solution. Can somebody let me know if I can make any optimizations in it.

Here is my code :

```
/*Priyansh Agarwal*/
#include<bits/stdc++.h>
using namespace std;
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
typedef long long ll;
void helper(int value,int *arr,int *arr1,int counter)
{
int i=0;
while(i<counter && arr[i]<=value)
{
int it=arr1[i]+1;
if(value%arr[i]==0)
{
int value1=pow(arr[i],it);
int temp=value;
while(true)
{
if(temp% value1==0)
{
arr1[i]++;
value1=value1*arr[i];
}
else
break;
}
}
i++;
}
}
ll helper1(int value,int*arr,int *arr1,int counter) //this function finds out the multiplication factor for the corresponding number passed to it);
{
int i=0;
ll sum=1;
while(i<counter && arr[i]<=value)
{
int it=arr1[i]+1;
if(value%arr[i]==0)
{
int value1=pow(arr[i],it);
int temp=value;
// ll sum=1;
while(true)
{
if(temp% value1==0)
{
value1=value1*arr[i];
sum=sum*ll(arr[i]);
}
else
break;
}
}
i++;
}
return sum;
}
int main()
{
fastio();
freopen("Input.txt","r",stdin);
freopen("Output.txt","w",stdout);
bool *sieve=new bool[10000+1]();
sieve[0]=true;
sieve[1]=true;
sieve[2]=false;
int *arr=new int[1230];
int *arr1=new int[1230]();
int counter=0;
for(int i=2;i<=10000;i++)
{
if(sieve[i]==false)
{
arr[counter++]=i;
for(int j=2*i;j<=10000;j+=i)
sieve[j]=true;
}
}
int t;
cin>>t;
while(t--)
{
int n,m;
for(int i=0;i<1230;i++)
arr1[i]=0;
cin>>n>>m;
int *arr3=new int[n];
bool *arr4=new bool[m+1]();
for(int i=0;i<n;i++)
{
cin>>arr3[i];
arr4[arr3[i]]=true;
helper(arr3[i],arr,arr1,counter);
}
int i=0;
// while(i<counter && arr[i]<=m)
// {
// cout<<arr[i]<<" "<<arr1[i]<<endl;
// i++;
// }
// cout<<"asdf"<<endl;
int max1=1;
ll max2=1;
for(int i=2;i<=m;i++)
{
if(arr4[i])
continue;
ll max3=helper1(i,arr,arr1,counter);
// cout<<i<<" "<<max3<<endl;
if(max3>max2)
{
max2=max3;
max1=i;
}
}
cout<<max1<<endl;
}
return 0;
}
```