Minimum cost of ropes to join

//for eg. 2 3 4 6
//let's find answer 2+3 ans=5,new array=[5,4,6] new_sum=5+4 or 5 is actually(2+3) so ans=2+3+2+3+4,new array=[9,6] new_sum=15
//9 is actually 2+3+4 so new sum=(((2+3)+2+3+4)+2+3+4+6) so the element we chose in first time is repeating so why not choose
//small element early for less repeating of elements so used min_heap
class Solution
{
    public:
    //Function to return the minimum cost of connecting the ropes.
    long long minCost(long long arr[], long long n) {
        priority_queue<long long int,vector <long long int>,greater<long long int>> pq;
        for(long long int i=0;i<n;i++) pq.push(arr[i]);
        long long int res=0;
        while(pq.size()>1){
            long long int t1=pq.top();pq.pop();
            long long int t2=pq.top();pq.pop();
            res+=(t1+t2);
            pq.push(t1+t2);
        }
        return res;
    }
};