Codeforces Round #321 (Div. 2)B. Kefa and Company

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 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 test(s)

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.

题目大意：n个朋友，分别有金钱值和友情值，邀请其中一些来聚会，要求来聚会的每个人金钱值低于d，并且友情值总量最大，求最大的友情值。

解题思路：贪心。先排序，金钱值递增，金钱值等价时友情值递减。枚举第一个朋友，再二分查找与他金钱值差异最大的且满足条件的朋友。注意：每次找到一组朋友后，累加友情值会超时，可以先预处理一下。

代码如下：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define ll long long
using namespace std;
const int maxn=100005;
ll sum[maxn];
struct frd
{
	ll x,y;
}fd[maxn];
ll cmp(frd u,frd v)
{
	if(u.x!=v.x)
		return u.x<v.x;
	else
		return u.y>v.y;
}
int main(void)
{
	ll n,d,ans=0,x,y;
	scanf("%I64d%I64d",&n,&d);
	for(ll i=0;i<n;i++)
	{
		scanf("%I64d%I64d",&fd[i].x,&fd[i].y);
	}
	sort(fd,fd+n,cmp);
	for(ll i=0;i<n;i++)
	{
		if(!i)
			sum[i]=fd[i].y;
		else
			sum[i]=sum[i-1]+fd[i].y;
	}
	for(ll i=0;i<n;i++)
	{
		ll cur=0,md=i;
		ll hg=n-1,lw=i,mid;
		while(hg>=lw)
		{
			mid=(hg+lw)/2;
			if(fd[mid].x-fd[i].x>=d)
				hg=mid-1;
			else
			{
				lw=mid+1;
				md=mid;
			}
		}
		cur=sum[md]-sum[i]+fd[i].y;
		ans=max(ans,cur);
	}
	printf("%I64d\n",ans );
}