Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn‘t want any friend to feel poor compared to somebody else in the company (Kefa doesn‘t count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1?≤?n?≤?105, 
) — the number of Kefa‘s friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa‘s friends, the (i?+?1)-th line contains the description of the i-th friend of type mi, si (0?≤?mi,?si?≤?109) — the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Example
Input
4 575 50 100150 2075 1
Output
100
Input
5 1000 711 3299 1046 887 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends.
先排序再利用尺取法，不断更新答案
1 #include<iostream>
2 #include<algorithm>
3 #include<cstdio>
4 #include<string>
5 using namespace std;
6
7 typedef long long ll;
8 struct node{
9 int mo;
10 int fr;
11 }N[100005];
12
13 bool operator <(const node& a,const node& b)
14 { return a.mo<b.mo;
15 }
16
17 int main()
18 { int n,d;
19 while(~scanf("%d%d",&n,&d)){
20 for(int i=1;i<=n;i++) scanf("%d%d",&N[i].mo,&N[i].fr);
21 sort(N+1,N+n+1);
22 int l=1,r=1;
23 ll ans=0,sum=0,temp=0;
24 while(r<=n){
25 sum=N[r].mo-N[l].mo;
26 temp+=N[r].fr;
27 while(sum>=d){ //一定是大于等于！！！
28 temp-=N[l].fr;
29 l++;
30 sum=N[r].mo-N[l].mo;
31 }
32 r++;
33 ans=max(ans,temp);
34 }
35 printf("%lld\n",ans);
36 }
37 return 0;
38 }