> #include <cstdio>
> #include <cstring>
> #include <algorithm>
> #include <iostream>
> #include <vector>
> #include <queue>
> #include <cmath>
> #include <set>
> using namespace std;
> 
> #define N 10005
> #define inf 999999999
> 
> __int64 max(__int64 a,__int64 b){return a>b?a:b;}
> __int64 min(__int64 a,__int64 b){return a<b?a:b;}
> 
> struct node{
> 	__int64 x, y;
> }a[N];
> 
> int n, d;
> main()
> {
> 	int i, j, k;
> 	int maxh, minh;
> 	while(scanf("%d %d",&n,&d)==2){
> 		maxh=-1;
> 		for(i=0;i<n;i++){
> 			scanf("%I64d %I64d",&a[i].x,&a[i].y);
> 			maxh=max(maxh,a[i].x);
> 			minh=min(minh,a[i].x);
> 		}
> 		maxh=(maxh-minh)/d;
> 		__int64 ans=a[0].x*a[0].y;
> 		for(i=1;i<n;i++){
> 			int v=-1;
> 			for(j=i-1;j>=0;j--){
> 				if(i-j>maxh) break;      //起码需要往前找maxh个，再往前面找就没有意义了
> 				int aa=(a[i].x-a[j].x)/(i-j);//初略估计时间复杂度最多为5000*4999/2+5000^2<5000W   不会超时 
> 				if(v<aa&&d<aa){
> 					v=aa;
> 					k=j;
> 				}
> 			}
> 			if(v!=-1) ans+=a[k].x*a[i].y+a[i].y*(__int64)(d*(i-k));
> 			else ans+=a[i].x*a[i].y;
> 		}
> 		printf("%I64d\n",ans);
> 	}
> }