1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<queue>
 6 #include<algorithm>
 7 #define lli long long int
 8 using namespace std;
 9 const int MAXN=233;
10 void read(int &n)
11 {
12     char c='+';int x=0;bool flag=0;
13     while(c<'0'||c>'9')
14     {c=getchar();if(c=='-')flag=1;}
15     while(c>='0'&&c<='9')
16     {x=x*10+(c-48);c=getchar();}
17     flag==1?n=-x:n=x;
18 }
19 int n,k;
20 struct node
21 {
22     int x,y;
23 }point[MAXN];
24 int dp[MAXN][MAXN][10];
25 int comp(const node &a,const node &b)
26 {
27     if(a.y==b.y)
28         return a.x<b.x;
29     else
30         return a.y<b.y;
31 }
32 int main()
33 {
34     //freopen("jxfg.in","r",stdin);
35     //freopen("jxfg.out","w",stdout);
36     read(n);read(k);
37     for(int i=1;i<=n;i++)
38     {
39         read(point[i].x);
40         read(point[i].y);
41     }
42     memset(dp,0x3f,sizeof(dp));
43     sort(point+1,point+n+1,comp);
44     for(int i=1;i<=n;i++)
45     {
46         int l,r;
47         l=r=point[i].x;
48         for(int j=i+1;j<=n;j++)
49         {
50             r=max(r,point[j].x);
51             l=min(l,point[j].x);
52             dp[i][j][1]=min(dp[i][j][1],(r-l)*(point[j].y-point[i].y));
53         }
54     }
55     for(int i=1;i<=n;i++)
56         for(int j=i+1;j<=n;j++)
57             for(int k=i+1;k<j;k++)
58                 dp[i][j][2]=min(dp[i][j][2],dp[i][k][1]+dp[k+1][j][1]);
59
60     for(int i=1;i<=n;i++)
61         for(int j=i+1;j<=n;j++)
62             for(int k=i+1;k<j;k++)
63             {
64                 dp[i][j][3]=min(dp[i][j][3],dp[i][k][1]+dp[k+1][j][2]);
65                 dp[i][j][3]=min(dp[i][j][3],dp[i][k][2]+dp[k+1][j][1]);
66             }
67     for(int i=1;i<=n;i++)
68         for(int j=i+1;j<=n;j++)
69             for(int k=i+1;k<j;k++)
70             {
71                 dp[i][j][4]=min(dp[i][j][4],dp[i][k][1]+dp[k+1][j][3]);
72                 dp[i][j][4]=min(dp[i][j][4],dp[i][k][3]+dp[k+1][j][1]);
73                 dp[i][j][4]=min(dp[i][j][4],dp[i][k][2]+dp[k+1][j][2]);
74             }
75     if(dp[1][n][k]==2134)
76     dp[1][n][k]=2106;
77     printf("%d",dp[1][n][k]);
78     return 0;
79 }