B. Kefa and Company

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Sample test(s)

input

4 5
75 5
0 100
150 20
75 1

output

100

input

5 1000 7
11 32
99 10
46 8
87 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.

解题说明：此题是求某个区间范围内的数，并让该数对应的值和最大。可以采用先排序，然后遍历的做法，在遍历过程中统计出每个区间中的最大值，最后得到一个全局的最大值。

#include<stdio.h>
#include <string.h>
#include<iostream>
#include<algorithm>
using namespace std;

int n,m,a,b,x,d;
int main()
{
cin>>n>>d;
pair<int,int>p[100005];
for(int i=0;i<n;++i)
{
cin>>p[i].first>>p[i].second;
}
sort(p,p+n);
long long ans=0,cur=0,st=0;
for(int i=0;i<n;)
{
if(p[i].first-p[st].first>=d)
{
cur-=p[st].second;
++st;
}
else
{
cur+=p[i].second;
++i;
}
ans=max(ans,cur);
}
cout<<ans<<endl;
return 0;
}