1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 typedef long long LL;
 7 const int N=5e4+5;
 8 const int INF=0x3f3f3f3f;
 9
10 int head,tail;
11 LL dp[N][105],q[N];
12
13 struct node{
14     LL w,h;
15     bool operator <(const node &b)const{
16         return w==b.w?h>b.h:w>b.w;
17     }
18 }a[N],tmp[N];
19
20 LL getUP(int m,int k,int j){
21     return dp[k][j-1]-dp[m][j-1];
22 }
23
24 LL getDOWN(int m,int k){
25     return a[m+1].h-a[k+1].h;
26 }
27
28 LL getDP(int i,int j,int k){
29     return dp[k][j-1]+a[k+1].h*a[i].w;
30 }
31
32 int main(){
33     int n,m;
34     while(scanf("%d%d",&n,&m)!=EOF){
35         for(int i=1;i<=n;i++){
36             scanf("%lld %lld",&tmp[i].w,&tmp[i].h);
37         }
38         sort(tmp+1,tmp+1+n);
39         int cnt=1;
40         a[1].h=tmp[1].h,a[1].w=tmp[1].w;
41         //删除会被完全覆盖的矩形
42         for(int i=1;i<=n;i++){
43             if(a[cnt].h<tmp[i].h){
44                 cnt++;
45                 a[cnt].h=tmp[i].h;
46                 a[cnt].w=tmp[i].w;
47             }
48         }
49         reverse(a+1,a+1+cnt);
50         for(int i=1;i<=cnt;i++){
51             dp[i][1]=a[1].h*a[i].w;
52         }
53         LL ans=dp[cnt][1];
54
55         for(int j=2;j<=m;j++){
56             head=tail=0;
57             q[tail++]=j-1;
58             dp[cnt][j]=INF;
59             for(int i=j;i<=cnt;i++){
60                 while(head+1<tail&&getUP(q[head],q[head+1],j)<=a[i].w*getDOWN(q[head],q[head+1])){
61                     head++;
62                 }
63                 dp[i][j]=getDP(i,j,q[head]);
64                 while(head+1<tail&&getUP(q[tail-1],i,j)*getDOWN(q[tail-2],q[tail-1])<=getUP(q[tail-2],q[tail-1],j)*getDOWN(q[tail-1],i)){
65                     tail--;
66                 }
67                 q[tail++]=i;
68             }
69             //加一句特判(从超时变成146ms),当挖洞数增多,花费未降低的时候,说明再增加挖洞数也不会降低花费,直接结束DP
70             if(ans<=dp[cnt][j])
71                 break;
72             else
73                 ans=dp[cnt][j];
74         }
75         printf("%lld\n",ans);
76     }
77     return 0;
78 }