Points on Line

Time limit 1000 ms Memory limit 32768 kB

来源：
codeforces：div1-251-A Points on Line
vjudge：H-Points on Line

单调队列

Description

Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed d.

Note that the order of the points inside the group of three chosen points doesn't matter.

Input

The first line contains two integers: n and d (1 ≤ n ≤ 105; 1 ≤ d ≤ 109). The next line contains n integers x1, x2, ..., xn, their absolute value doesn't exceed 109 — the x-coordinates of the points that Petya has got.

It is guaranteed that the coordinates of the points in the input strictly increase.

Output

Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed d.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Example

Input

4 3
1 2 3 4

Output

4

Input

4 2
-3 -2 -1 0

Output

2

Input

5 19
1 10 20 30 50

Output

1

Note

In the first sample any group of three points meets our conditions.

In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.

In the third sample only one group does: {1, 10, 20}.

题意：

输入第一行包括两个数：n,d;
接下来输入n个数；
任意三个数，最大数与最小数之间的差<=d;
问这样的数有几组；

解题思路：

利用单调队列，将这n个数push（）入队列，将第一个数front()与最后一个数back()比较，如果他们之间的差小于d，则结果ans=Cn2，，就是n*（n-1）；在前n个数里选2个，如果他们之间的差大于d就将第一个元素弹出，直到可以为止
   最后记得将队列里的元素都弹出来！！！

完整代码：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
int main()
{
    queue<int>q;
    int n,d,m;
    while(scanf("%d%d",&n,&d)!=EOF)
    {
        ll ans=0,count=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&m);
            q.push(m);
            count++;
            while(q.back()-q.front()>d&&!q.empty())
             {
                 q.pop();
                 count--;
             }
            ans+=((count-1)*(count-2))/2;///为什么不可以用q.size(),,我不是很懂，我觉得都一样的呀！
        }
        while(q.size())
            q.pop();
        cout<<ans<<endl;
    }
    return 0;
}