CodeForces

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.

Examples

input

4 575 50 100150 2075 1

output

100

input

5 1000 711 3299 1046 887 54

output

111

题意：  小明有n个朋友  这里有一个m表示最大的贫富差距，，，每个朋友有他的  财富值 和 价值num  小明要邀请朋友们参加聚会  为了使 朋友们不尴尬 （因为贫富差距 ）不尴尬的条件是  参加聚会的朋友的财富值差距要小于m  并且在不尴尬的条件下使得  参加聚会的总价值最大  问总价值最大为多少。

思路： 尺取法求最大价值 ； 先将财富值由小到大排序  两个指针i（右指针）和j（左指针）  每次sum<m i++;  每次sum>= m  j++;

每次更新最大价值。

代码： 

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define N 100005using namespace std;struct node{long long mon;long long num;}a[N];long long n,m;bool cmp(node a,node b){return a.mon<b.mon;}int main(){int i,j;scanf("%lld %lld",&n,&m);for(i=1;i<=n;i++) scanf("%lld %lld",&a[i].mon,&a[i].num);sort(a+1,a+n+1,cmp);i=j=0;long long  maxx=-1;long long sum=0;while(i<=n){if(abs(a[i].mon-a[j].mon)<m){sum+=a[i].num;i++;}else{sum-=a[j].num;    j++;}maxx=max(maxx,sum);}printf("%lld\n",maxx);return 0;}