import java.util.*;
public class Main {
 
/*Cost[j,i]表示安排前i个邮局给前j个村庄使用
   * D[i,j]表示在第i个到第j个村庄中建立一个邮局得到的最短距离总和。
   * D[a,b]可以利用关系式d[a,a]=0, D[a,b]=D[a,b-1]+x[b]-x[(a+b)/2]表示*/
public static void main(String[] args) {
   // TODO Auto-generated method stub
   Scanner cin = new Scanner(System.in);
   int n, m;
   n = cin.nextInt();
   m = cin.nextInt();
   int[] x = new int[n + 1];
   int[][] d = new int[n + 2][n + 2];
   int[][] cost = new int[n + 2][n + 2];
   for (int i = 1; i <= n; i++)
    x[i] = cin.nextInt();
   for (int i = 1; i <= n; i++){ 
    for (int j = 1; j <= n; j++){
     if (i >= j)
      d[i][j] = 0;
     else
      d[i][j] = d[i][j - 1] + x[j] - x[(i + j) / 2]; 
    }
   }
   for (int i = 1; i <= n; i++)
    cost[i][1] = d[1][i];
   for (int i = 2; i <= m; i++ ){ 
    for (int j = i + 1; j <= n; j++){
     int tmp = Integer.MAX_VALUE;
     for (int k = i - 1; k <= j; k++){
      if ( tmp > cost[k][i - 1] + d[k + 1][j] )
       tmp = cost[k][i - 1] + d[k + 1][j]; // Cost[n,m] = min{Cost[i,m-1]+D[i+1,n]}，i=m-1,m,…,n-1。
     }
     cost[j][i] = tmp;
    }
   }
   System.out.println(cost[n][m]);
}
 
}