B

B - Kefa and Company

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 580B

Description

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.

Sample Input

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

Hint

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.

题目大意：Kefa要请朋友吃饭，他有n个朋友，这些朋友都两个特征：1.身上所带钱数2.对Kefa的友谊值

如果这些朋友中有人所带钱数比这个朋友所带钱所多与超过d元（包括d），那么这朋友会觉得自己可怜，Kefa

不想让自己的朋友感到可怜，但他又想获得高得友谊值，问Kefa能获得的最高的友谊值是多少

解题：

将朋友所带的钱数和友谊值排序，按钱数从小到大排然后进行判断处理

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>

using namespace std;

const int N = 100010;
typedef long long ll;

struct st
{
    ll m, s;
}node[N];

int cmp(const void *a, const void *b)
{
    st *s1 = (st *)a, *s2 = (st *)b;
    if(s1->m != s2->m)
        return s1->m - s2->m;
    return s1->s - s2->s;
}

int main()
{
    int n, i;
    ll d, sum;
    while(~scanf("%d%I64d", &n, &d))
    {
        for(i = 0 ; i < n ; i++)
            scanf("%I64d%I64d", &node[i].m, &node[i].s);
        qsort(node, n, sizeof(node[0]), cmp);
        sum = i = 0;
        int j = 0;
        ll Max = 0;
        while(i < n)
        {
            if(node[i].m - node[j].m >= d)
            {
                sum -= node[j].s;
                j++;
            }
            else
            {
                sum += node[i].s;
                i++;
            }
            Max = max(Max, sum);
        }
        printf("%I64d
", Max);
    }
    return 0;
}