// Abordare brute force O(n^2)

#include <iostream>
#include <fstream>
#define nmax 5000001
#define nr 100000000
using namespace std;

ifstream f("deque.in");
ofstream g("deque.out");

int v[nmax],n,k,i,j,minim,cnt;
long long suma;

int main()
{
f>>n>>k;
for(i=1;i<=n;i++)
    f>>v[i];

for(i=1;i<=n-k+1;i++)
{
    minim=nr;
     for(j=0;j<k;j++)
        if(v[i+j]<minim)
           minim=v[i+j];

     suma=suma+minim;
}
g<<suma;

    return 0;
}