codeforces-580B Kefa and Company(二分)



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 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.

这道题是二分法

先按朋友们的钱从小到大进行排序;

然后二分搜索(1~N)之间所能容纳最大金钱差的人;

完成后得到所对应的友情值最大的就是最终答案;

#include<stdio.h>
#include<algorithm>
using namespace std;
long long  b[100005];
int n,d,i,l,r,mid;
long long maxx,ans;
struct p
{

int m;
int s;

}a[100005];
bool camp(p aa,p bb)
{

return aa.m<bb.m;

}
int main()
{
scanf("%d%d",&n,&d);
for (i=0;i<n;i++)
{
scanf("%d%d",&a[i].m,&a[i].s);
}
sort(a,a+n,camp);
b[0]=0;
maxx=-1;
for (i=1;i<=n;i++)
{

b[i]=a[i-1].s+b[i-1];

}
for (i=0;i<n;i++)
{
l=i;
r=n;
while (r-l>1)
{
mid=(l+r)/2;
if (a[mid].m-a[i].m<d) l=mid;
else r=mid;
}
ans=b[l+1]-b[i];
if (ans>maxx)  maxx=ans;

}
printf("%I64d",maxx);

}