tj

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

const int N = 2e5 + 10;
int a[N];
int n,c;

int main(){
	cin >> n >> c;
	for(int i = 1;i <= n;i++){
		cin >> a[i];
	}
	sort(a + 1,a + 1 + n); //不一定有序,先排序 

    long long ans = 0;
	for(int i = 1;i <= n;i++){
		int b = a[i];
		int A = b + c; //计算 A 
		//找到 A 的个数 
		int p1 = upper_bound(a + 1,a + 1 + n,A) - a;
		int p2 = lower_bound(a + 1,a + 1 + n,A) - a;
		ans += p1 - p2;
	}
	cout << ans << endl;
	return 0;
}