#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")

using namespace std;
#include <atcoder/all>
using namespace atcoder;

#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
typedef long long ll;
#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define REP(i,num,n) for(int i=num, i##_len=(n); i<i##_len; ++i)
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
#define print(x) (cout << (x) << endl)
#define pb push_back
#define mp make_pair
#define sz(x) int(x.size())
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
#define fs first
#define sc second
template<typename T> vector<T> make_vec(size_t n, T a) { return vector<T>(n, a); }
template<typename... Ts> auto make_vec(size_t n, Ts... ts) { return vector<decltype(make_vec(ts...))>(n, make_vec(ts...)); }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &v) { for (auto &e : v) os << e << ' '; return os; }
// 多倍長
#include <boost/rational.hpp>
#include <boost/multiprecision/cpp_int.hpp>
using boost::rational;
using boost::multiprecision::cpp_int;

const double PI = acos(-1);
const int INF = 1e9;
const ll LLINF = 1e16;
const ll MOD = 1e9+7;
template<class T> inline void add(T &a, T b){a = ((a+b) % MOD + MOD) % MOD;};

 

// 指定した文字cが文字列に何文字入ってるか
ll stringcount(string s, char c) {
    return count(s.cbegin(), s.cend(), c);
}


// 割り算の切り上げ処理
template<typename T>
T rp(T a, T b) {
  return (a + b - 1) / b;
}

// 桁和
template<typename T>
T digsum(T n) {
    T res = 0;
    while(n > 0) {
        res += n%10;
        n /= 10;
    }
    return res;
}

// 回文判定
bool kai (string s) {
    string t{s};
    reverse(begin(t), end(t));
    return s == t;
}

using pint = pair<int, int>;
using pll = pair<ll, ll>;

ll dp[500000];
int main(void)
{
    cout.tie(0);
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    ll n,k;
    cin >> n >> k;
    ll ans = 1;
    ll prev = 0;
    rep(i, n){
        ll a;
        cin >> a;
        if(i != 0){
            if(abs(prev - a) <= k) ans++;
        }
        prev = a;
    }
    print(ans);
}
