## Algorithm

C. 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 ะก++. It is preferred to use the cincout streams or the %I64d specifier.

Examples
input
Copy
`4 31 2 3 4`
output
Copy
`4`
input
Copy
`4 2-3 -2 -1 0`
output
Copy
`2`
input
Copy
`5 191 10 20 30 50`
output
Copy
`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}.

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include<bits/stdc++.h>
using namespace std;

#define ll long long
// #define endl "\n"
#define debug(n) cout<<(n)<<endl;
const ll INF = 2e18 + 99;

int binary_search(ll arr[], int low, int high, ll target){
while(low < high){
int mid = (low + high)/2;
if(arr[mid] > target){
high = mid -1;
}
else{
if(mid == high){
return mid;
}
if(arr[mid+1] > target){
return mid;
}
low = mid +1;
}
}
return low;
}

int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);

ll n, d;
cin>>n>>d;
ll arr[n];
for(int i = 0; i < n; i++){
cin>>arr[i];
}
ll ans = 0;
for(int i = 0; i < n; i++){
ll j = binary_search(arr, i, n-1, arr[i]+d);
if(j - i > 1){
ll cnt = j - i + 1;
ans += ((cnt-2)*(cnt-1))/2;
}
}
cout<<ans<<endl;
}``````
Copy The Code &

Input

cmd
4 3
1 2 3 4

Output

cmd
4

## Demonstration

Codeforcess Solution 252-C C. Points on Line ,C++, Java, Js and Python,252-C,Codeforcess Solution