/*

 线段树优化dp

 dp[i]表示前i个数的最长为多少，则dp[i]=max(dp[j]+1) abs(a[i]-a[j])<=d

 复杂度为O(n ^ 2)

 利用线段树优化，线段树保存区间最大值。离散化后便可求出，还要注意 对于叶子节点保存的即为dp的值，每次更改即可，开始一直累加。。。。。

 */

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <cmath>

 #include <algorithm>

 using namespace std;

 #define lson l,m,rt<<1

 #define rson m + 1, r, rt<<1|1

 const int maxn = 1e5 + ;

 int s[maxn];

 int n, d;

 int san[maxn], tot;

 int sum[maxn << ];

 void pushUp(int rt){

     sum[rt] = max(sum[rt<<], sum[rt<<|]);

 }

 void update(int pos, int c, int l, int r, int rt){

     if (l == r){

         sum[rt] = c;//注意

         return ;

     }

     int m = (l + r) >> ;

     if (pos <= m) update(pos, c, lson);

     else update(pos, c, rson);

     pushUp(rt);

 }

 int query(int L, int R, int l, int r, int rt){

     if (L <= l && R >= r){

         return sum[rt];

     }

     int m = (l + r) >> ;

     int ret = ;

     if (L <= m) ret = query(L, R, lson);

     if (R > m) ret = max(ret, query(L, R, rson));

     return ret;

 }

 int main(){

     while (~scanf("%d%d", &n, &d)){

         tot = ;

         for (int  i = ; i <= n; ++i){

             scanf("%d", &s[i]);

             san[tot++] = s[i];

         }

         sort(san, san + tot);

         tot = unique(san, san + tot) - san;

         memset(sum, , sizeof(sum));

         int ans = ;

         for (int i = ; i <= n; ++i){

             int pos = lower_bound(san, san + tot, s[i]) - san + ;

             int l = lower_bound(san, san + tot, s[i] - d) - san + ;

             int r = upper_bound(san, san + tot, s[i] + d) - san;

             int que = query(l, r, , tot, ) + ;

             //cout << " l = " << l << " r = " << r << endl;

             ans = max(ans, que);

             //cout << " ans = " << ans << endl;

             update(pos, que, , tot, );

         }

         printf("%d\n", ans);

     }

     return ;

 }