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

int main(){
    ifstream fin("deque.in");
    ofstream fout("deque.out");
    unsigned int n, k, i, j;
    fin>>n>>k;
    long long val, sum = 0, min = INT64_MAX, a[n];
    deque<long long> deq;
    for(i = 0; i < n; i++)
        fin>>a[i];
    
    for(i = 0; i <= n; i++){
        while(!deq.empty() && a[i] <= a[deq.back()])
            deq.pop_back();
        
        if(!deq.empty() && i-k-1 == deq.front())
            deq.pop_front();

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

        deq.push_back(i);
    }
    fout<<sum;

    return 0;
}