CodeForces - 580B Shell Game

題目網址： http://codeforces.com/problemset/problem/777/A

Kefa and Company

 

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 m_i, s_i (0 ≤ m_i, s_i ≤ 10⁹) — the amount of money and the friendship factor, respectively.

Output

Print the maximum total friendship factir that can be reached.

Example

Input

4 5
75 5
0 100
150 20
75 1

Output

100

Input

5 100
0 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.

題解：在比較度不超過一個數值的情況下，友好值達到最高。典型的取尺法，先根據錢的多少從小到大排序，然後從前往後跑一遍。

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
    int m,s;
}no[100005];
bool cmp(node x,node y){
    return x.m<y.m;
}
int main()
{
    int n,d,i,j;
    long long int maxn;
    long long int sum1,sum2;
    while(~scanf("%d%d",&n,&d)){
        maxn=0;
        for(i=0;i<n;i++){
            scanf("%d%d",&no[i].m,&no[i].s);
        }
        sort(no,no+n,cmp);
        sum1=0;sum2=0;
        j=0;
        for(i=0;i<n;){
            if(no[i].m-no[j].m>=d){
                sum2-=no[j].s;
                j++;
            }
            else{
                sum2+=no[i].s;
                i++;
            }
            maxn=max(maxn,sum2);
        }
        printf("%lld\n",maxn);
    }
    return 0;
}