【67.23%】【CF 716A】Crazy Computer

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

ZS the Coder is coding on a crazy computer. If you don't type in a word for a c consecutive seconds, everything you typed disappear!

More formally, if you typed a word at second a and then the next word at second b, then if b - a ≤ c, just the new word is appended to other words on the screen. If b - a > c, then everything on the screen disappears and after that the word you have typed appears on the screen.

For example, if c = 5 and you typed words at seconds 1, 3, 8, 14, 19, 20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.

You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything.

Input

The first line contains two integers n and c (1 ≤ n ≤ 100 000, 1 ≤ c ≤ 109) — the number of words ZS the Coder typed and the crazy computer delay respectively.

The next line contains n integers t1, t2, ..., tn (1 ≤ t1 < t2 < ... < tn ≤ 109), where ti denotes the second when ZS the Coder typed the i-th word.

Output

Print a single positive integer, the number of words that remain on the screen after all n words was typed, in other words, at the secondtn.

Examples

input

6 5
1 3 8 14 19 20

output

3

input

6 1
1 3 5 7 9 10

output

2

Note

The first sample is already explained in the problem statement.

For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1.

【题解】

和前一个对比。如果差值小于等于就递增数目。

如果差值大于就把数目改成1。。

就是道简单的模拟题。

注意第一个数字不一定全是1.所以不要把第一个数字的前一个数字想成是0.直接记录数目为1.然后从第2个开始处理就好了。

【代码】

#include <cstdio>

int n, c;

int main()
{
	//freopen("F:\rush.txt", "r", stdin);
	scanf("%d%d", &n, &c);
	int pret,num = 1;
	scanf("%d", &pret);
	for (int i = 2; i <= n; i++)
	{
		int x;
		scanf("%d", &x);
		int temp = x - pret;
		if (temp <= c)
			num++;
		else
			num = 1;
		pret = x;
	}
	printf("%d
", num);
	return 0;
}