// Monotonic deque implementation, O(n) time complexity
#include <fstream>
#include <deque>
using namespace std;

int main(){
    ifstream fin("deque.in");
    ofstream fout("deque.out");
    int n, k, i;
    long long sum = 0;
    fin>>n>>k;
    int a[n+1];
    deque<int> deq;

    for(i = 1; i <= n; i++)
        fin>>a[i];
    
    for(i = 1; i <= n; i++){
        while(!deq.empty() && a[i] <= a[deq.back()])
            deq.pop_back();

        deq.push_back(i);

        if(i-k >= 0)
            sum += a[deq.front()];

        if(i-deq.front()+1 >= k)
            deq.pop_front();
    }
    fout<<sum;

    return 0;
}