1 #include<stdio.h>
 2 #include<iostream>
 3 #include<string.h>
 4 #include<vector>
 5 #include<math.h>
 6 #include<algorithm>
 7 #include<set>
 8 #include<string.h>
 9 #include<string>
10 #include<map>
11 #include<queue>
12 using namespace std;
13 #define N 100008
14 #define ll long long
15 #define ull unsigned long long
16 #define inf 0x3f3f3f3f
17 struct node{
18     ll dis,num;
19 }p[N];
20 ll num[N],d[N];//d[i]表示p[i]到p[i-1]的距离
21 ll s4[N];
22 
23 int main(){
24     int n;
25     ll L;
26     scanf("%d%I64d",&n,&L);
27     int x,y;
28     num[0]=0ll;
29     for(int i=1;i<=n;i++){
30             num[i]=0ll;
31         scanf("%I64d%I64d",&p[i].dis,&p[i].num);
32         num[i]=num[i-1]+p[i].num;
33         if(i==1)d[i]=0;
34         else d[i]=p[i].dis-p[i-1].dis;
35     }
36     p[n+1].dis=L;p[n+1].num=0;
37     if(n<=3){
38         printf("0\n");
39     }
40     else{
41         s4[n+1]=0;
42         for(int i=n;i>=1;i--){//从i点到n+1点所有人需要走到s4的距离
43             s4[i]=s4[i+1]+ (p[n+1].dis-p[i].dis)*p[i].num;
44         }
45 
46         ll ans=(ll)inf*1000;
47         ll s1=0,s2=0,s3=0;
48         for(int i=1;i<=n-2;i++){//暴力枚举三个点的位置
49             s1+= num[i-1]*d[i];
50             s2=0;
51             if(s1>=ans)continue;
52             for(int j=i+1;j<=n-1;j++){
53                     s2+= ( num[j-1]-num[i])*d[j];
54                     s3=0;
55                     if(s1+s2>=ans)continue;
56                 for(int k=j+1;k<=n;k++){
57                     s3+=(num[k-1]-num[j])*d[k];
58                     ll s=s4[k+1];//表示最后一段的结果
59                     if(ans>s1+s2+s3+s){
60                        // printf("%d %d %d %d\n",i,j,k,n+1);输出新区间
61                         ans=s1+s2+s3+s;
62                     }
63                 }
64             }
65         }
66         printf("%I64d\n",ans);
67     }
68 
69     return 0;
70 }