A. Points on Line time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

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, ..., x_n, 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.

Sample test(s) 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}.

分析；n個點，距離d，找出集合中每三點最遠和遠近點距離不超過d的集合。

首先確定起點，然後二分確定最大點，從而確定了一個範圍，第二小點就在最大點與最小點間選擇。

這題和山大的那道三角形的方法是一樣的。

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
long long num[100005],n,d;
int main()
{
    long long i,ans=0;
    cin>>n>>d;
    for (i = 0; i < n; i ++)
        cin>>num[i];
    for (i = 0; i < n - 2 ; i ++)
    {
    long long mid,start=i,a=i+2,b=n-1;
    while (a < b)
    {
        mid = (a + b) / 2;
        if (num[mid]  - num[start] <= d)
            a = mid + 1;
        else
            b = mid;
    }
    mid = (a + b) / 2;
    if (num[mid] - num[start] > d)
        mid --;
        if (mid - i < 2) continue;
        ans += (mid- i) * (mid- i - 1) / 2;
    }
    cout<<ans<<endl;
    return 0;
}