CodeForces 251A 二分

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 %I64dspecifier.

Sample Input

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

Hint

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}.

题意：n个点，距离d，找出集合中每三点最远和远近点距离不超过d的集合。

思路：选定起点，然后用二分找到上界，在求出个数，复杂度为nlogn

代码：

#include <stdio.h>
#include <string.h>

long long n, d;
long long i, j;
long long num[100005];
long long ans = 0;

long long find(long long  start, long long i, long long j) {
	long long mid;
	while (i < j) {
		mid = (i + j) / 2;
		if (num[mid]  - num[start] <= d) {
			i = mid + 1;
		}
		else {
			j = mid;
		}
	}
	mid = (i + j) / 2;
	if (num[mid] - num[start] > d)
		mid --;
	return mid;
}


int main() {
	scanf("%lld%lld", &n, &d);
	for (i = 0; i < n; i ++)
		scanf("%lld", &num[i]);
	for (i = 0; i < n - 2 ; i ++) {
		int j = find(i, i + 2, n - 1);
		if (j - i < 2) continue;
		ans += (j - i) * (j - i - 1) / 2;
	}
	printf("%lld\n", ans);
	return 0;
}