#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define f first
#define s second
#define pii pair<int,int>
#define is insert
const int INF = 2e9;
const int MOD = 1e9 + 7;
const int MAXN = 2e5 + 1;
//store test cases here
/*
*/
int n,x;
vector<int> a;
vector<int> d;
vector<int> r;
int f[MAXN];
int b[MAXN];
int l[MAXN];
vector<pii> sor;
int t[4*MAXN];
void build(int curr, int left, int right) {
if (left == right ) {
t[curr] = 0;
} else {
int mid = (right + left) / 2;
//an edge's two children are the indices 2i && 2i+1
//in this sum segtree, an edge's value is the values from it's left to right indices
build(curr * 2, left, mid);
build(curr*2 + 1, mid + 1, right);
t[curr] = max(t[curr*2],t[curr*2+1]);
}
//cout << "tl: " << left << " tr: " << right << " val: " << t[curr] << "\n";
}
//tl-tr is the current segment, and l-r is the desired segment
int sum(int curr, int tl, int tr, int left, int right) {
if (left > right) {
return 0;
}
if (left == tl && right == tr) {
return t[curr];
}
int mid = (tl + tr) / 2;
//taking min of r and mid, because mid might be greater than r, same for max
return max(sum(curr*2, tl, mid, left, min(right,mid)), sum(curr*2+1, mid +1, tr, max(left,mid+1), right));
}
void update(int curr, int tl, int tr, int pos, int val) {
if (tl == tr) {
t[curr] = val;
} else {
int mid = (tl+tr) / 2;
//see which segment the index falls in, and change it
if (pos <= mid) {
update(curr*2, tl, mid, pos, val);
} else {
update(curr*2+1, mid+1, tr, pos, val);
}
//update the curr segmment to the new val once everything below it has been updated properly
t[curr] = max(t[curr*2],t[curr*2+1]);
}
//cout << "tl: " << tl << " tr: " << tr << " val: " << t[curr] << "\n";
}
int bsearch(int a) {
int lo = 0;
int hi = n-1;
while (lo != hi) {
int mid = (lo+hi)/2;
if (r[mid] < a) {
hi = mid;
} else {
lo = mid+1;
}
//cout << lo << "\n";
}
return lo;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n >> x;
for (int i = 0; i < n; i++) {
int m;
cin >> m;
a.pb(m);
sor.pb(mp(m,i));
d.pb(INF);
r.pb(-1*INF);
}
d.pb(INF);
d[0] = -1*INF;
r.pb(-1*INF);
r[0] = INF;
for (int i = 0; i < n; i++) {
int ind = lower_bound(d.begin(),d.end(),a[i]+1) - d.begin();
if (d[ind-1] < a[i] && a[i] < d[ind]) {
d[ind] = a[i];
f[i] = ind;
} else {
f[i] = 1;
}
//cout << "i: " << i << " val: " << f[i] << "\n";
}
for (int i = n-1; i > -1; i--) {
int ind = bsearch(a[i]-1);
if (r[ind] < a[i] && a[i] <= r[ind-1]) {
r[ind] = a[i]-1;
b[i] = ind;
} else {
b[i] = 1;
}
//cout << "i: " << i << " val: " << b[i] << "\n";
}
int ans = 0;
for (int i = 0; i <= n; i++) {
if (d[i] < INF) {
ans = i;
}
}
//empty segtree with size n, point update each time, query minimum in a range
//find first index that is geq and last index that is leq
//put elements in set afterwards, and map each element to an index in sorted segtree
map< pii, int> p;
sort(sor.begin(), sor.end());
for (int i = 0; i < n; i++) {
p.insert(mp(sor[i],i));
//cout << "(" << sor[i].first << "," << sor[i].second << ") i: " << i << "\n";
}
build(1,0,n-1);
set<pii> taken;
//insert beg
taken.insert(mp(INF, n));
taken.insert(mp(-1*INF, -1*INF));
taken.insert(mp(a[0], p[mp(a[0],0)]));
update(1,0,n-1,p[mp(a[0],0)],f[0]);
//cout << sum(1,0,n-1,5,6) << "\n";
for (int i = 1; i < n; i++) {
int lo = taken.lower_bound(mp(a[i],0))->second;
int hi = taken.lower_bound(mp(a[i]+x,0))->second;
//cout << "i:" << i << " a[i]:" << a[i] << " lo:" << lo << " hi:" << hi << " sum: " << sum(1,0,n-1,lo,hi-1) << "\n";
ans = max(ans,sum(1,0,n-1,lo,hi-1) + b[i]);
taken.insert(mp(a[i], p[mp(a[i],i)]));
update(1,0,n-1,p[mp(a[i],i)],f[i]);
//cout << "ind: " << p[mp(a[i],i)] << " f[i] " << f[i] << "\n";
}
cout << ans << "\n";
}
/* REMINDERS
* CHECK ARRAY BOUNDS, HOW BIG ARRAY HAS TO BE
* PLANNING!!!!!!!! Concrete plan before code
* IF CAN'T FIGURE ANYTHING OUT, MAKE TEN TEST CASES TO EVALUATE ALL TYPES OF SCENARIOS, THEN CONSTRUCT SOLUTION TO FIT IT
* IF CAN'T FIGURE ANYTHING OUT, MAKE TEN TEST CASES TO EVALUATE ALL TYPES OF SCENARIOS, THEN CONSTRUCT SOLUTION TO FIT IT
* IF CAN'T FIGURE ANYTHING OUT, MAKE TEN TEST CASES TO EVALUATE ALL TYPES OF SCENARIOS, THEN CONSTRUCT SOLUTION TO FIT IT
* NAIVE SOL FIRST TO CHECK AGAINST OPTIMIZED SOL
* MOD OUT EVERY STEP
* DON'T MAKE ASSUMPTIONS
* DON'T OVERCOMPLICATE
* CHECK INT VS LONG, IF YOU NEED TO STORE LARGE NUMBERS
* CHECK CONSTRAINTS, C <= N <= F...
* CHECK SPECIAL CASES, N = 1...
* TO TEST TLE/MLE, PLUG IN MAX VALS ALLOWED AND SEE WHAT HAPPENS
* ALSO CALCULATE BIG-O, OVERALL TIME COMPLEXITY
* IF ALL ELSE FAILS, DO CASEWORK
* compile with "g++ -std=c++11 filename.cpp" if using auto keyword
*/
