Given a sequence of integers ‘a’, a triplet (a[i], a[j], a[k]) is beautiful if:

1) i < j < k

2) a[j] -a[i] = a[k]-a[j] = d

Given an increasing sequence of integers and the value of ‘d’, count the number of beautiful triplets in the sequence.

**Input:**
The first line 2 space-separated integers 'n' and 'd', the length of the sequence and the beautiful difference.
The second line 'n' space-separated integers arr[i].
**Output:**
Print a single line denoting the number of beautiful triplets in the sequence.

**Explanation**:**Sample Input**

STDIN Function

7 3 arr[] size n = 7, d = 3

1 2 4 5 7 8 10 arr = [1, 2, 4, 5, 7, 8, 10]

**Sample Output**

3

There are many possible triplets (arr[i], arr[j], arr[k]), but our only beautiful triplets are (1,4,7), (4,7,10) and (2,5,8) by value, not index. Please see the equations below:

7-4 = 4-1 =3 =d

10-7 = 7-4 =3 =d

8-5 = 5-2 =3 =d

Recall that a beautiful triplet satisfies the following equivalence relation:

arr[j] – arr[i] = arr [k] – arr [j] = d where I<j<k.

#include <stdio.h> #include<stdlib.h> int main() { int str[100]; int n,d,s=0,i,j,k; scanf("%d %d",&n,&d); int *arr; arr=(int *)malloc(n*sizeof(int)); *arr=n; for( i=0;i<n;i++) {scanf("%d",&str[i]); } for( i=0;i<n;i++){ for( j=i+1;j<n;j++){ if(str[j]-str[i]!=d) continue; for( k=j+1;k<n;k++){ if(str[j]-str[i]==str[k]-str[j] && str[k]-str[j]==d)s++; } } } printf("%d",s); return 0; }

**INPUT_1:**

6 3

2 4 5 7 8 10

**OUTPUT:**

2

**INPUT_2:**

5 3

4 5 7 8 10

**OUTPUT:**

1

**ILLUSTRATION**