A. Points on Line

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, ..., 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 %I64dspecifier.

Examples

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}.

思路：因为给的序列是单调的，那么我么只要看当前点和头结点是否相差超过m，不超过的话sum+=C(i-l，2)；

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
using namespace std;
typedef long long LL;
LL ans[100005];
LL dp[100005];
int main(void)
{
    int n,m;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        int i,j;LL sum = 0;
        for(i = 0;i < n;i++)
        {
            scanf("%lld",&ans[i]);
        }
        if(n <= 2)
            printf("0
");
        else
        {
            int l = 0;
            for(i = 2;i < n;i++)
            {
                while(ans[i]-ans[l]>m)
                {
                    l++;
                }
                sum = sum + (LL)(i-l)*(LL)(i-l-1)/(LL)2;
            }printf("%lld
",sum);
        }
        
    }
    return 0;
}