//to live is to die
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long int ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<pair<int, int>> vpi;
typedef vector<pair<ll, ll>> vpl;
#define Clear(a, n) \
for (int i = 0; i <= n; i++) \
{ \
a[i] = 0; \
}
#define clearMat(a, n, m, d) \
for (int i = 0; i <= n; i++) \
{ \
for (int j = 0; j <= m; j++) \
a[i][j] = d; \
}
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define PB push_back
#define PF push_front
#define MP make_pair
#define F first
#define S second
#define rep(i, n) for (int i = 0; i < n; i++)
#define repe(i, j, n) for (int i = j; i < n; i++)
#define SQ(a) (a) * (a)
#define rep1(i, n) for (int i = 1; i <= n; i++)
#define Rrep(i, start, finish) for (int i = start; start >= finish; i--)
#define db(x) cerr << #x <<" "; _print(x); cerr << endl;
#define forn(i, Start, End, step) for (int i = Start; i <= End; i += step)
#define rforn(i, Start, End, step) for (int i = Start; i >= End; i -= step)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
// ll arr[SIZE];
/*
how to find n % mod ; n < 0?
x = (n+mod)%mod
if(x < 0) x += mod;
*/
void _print(int x)
{
cerr << x;
}
void _print(ll x)
{
cerr << x;
}
void _print(string x)
{
cerr << x;
}
void _print(char x)
{
cerr << x;
}
void _print(double x)
{
cerr << x;
}
void _print(ull x)
{
cerr << x;
}
void _print(vl x)
{
for(auto e : x)
{
cerr << e << " ";
}
cerr << "\n";
}
void print(vpi x)
{
for(auto e : x)
{
cerr << e.F << " " << e.S << "\n";
}
cerr << "\n";
}
void _print(vi x)
{
for(auto e : x)
{
cerr << e << " ";
}
cerr << "\n";
}
void _print(deque<ll>x)
{
for(auto e : x)
{
cerr << e << " ";
}
cerr << "\n";
}
//order_of_key(k): # of elements less than k (which is the index of x = k)
//find_by_order(k); iterator of the k-th element
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal <T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T> bool ckmin(T& a, const T& b)
{
return b<a?a=b,1:0;
}
template<class T> bool ckmax(T& a, const T& b)
{
return a<b?a=b,1:0;
}
template<typename T> istream& operator>>(istream& in, vector<T>& a)
{
for(auto &x : a) in >> x;
return in;
};
template<typename T> ostream& operator<<(ostream& out, vector<T>& a)
{
for(auto &x : a) out << x << ' ';
return out;
};
// priority_queue<data type , the container that would hold the values , greater<pair<int,int>>>
// greater means that we want the smallest value on top
// less means that we want the largest
// x ^ (n) mod m = ( (x mod m)^(n) ) mod m
char to_char(int num)
{
return (char)(num + '0');
}
ll const MAX = 1e18+1;
ll const oo = 1e18 + 1;
ll const INF = 1e9 + 10;
const ll MOD = 1e9 + 7;
ll const SIZE = 2e5 + 900;
//const int MAX_N = 100'005;
const int LOG = 20;
void solve()
{
ll N , M;
cin >> N >> M;
vl A(N), vec;
rep(i , N)
{
cin >> A[i];
ll val = (i + 1LL)*M - A[i];
if(val >= 0)
{
vec.PB(val);
}
//negative value will be changed always
}
vl dp((int)vec.size() + 1 , 1e18);
dp[0] = -1e18;
int LIS = 0;
for(ll& val : vec)
{
int L = upper_bound(all(dp) , val) - dp.begin();
if(dp[L - 1] <= val && val <= dp[L])
{
dp[L] = val;
LIS = max(LIS , L);
}
}
cout << N - LIS << "\n";
}
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
// freopen("cowjog.in" , "r" , stdin);
// freopen("cowjog.out", "w" , stdout);
int T = 1;
// cin >> T;
while(T--)
{
solve();
}
return 0;
}
/* stuff you should look for
* WRITE STUFF DOWN, ON PAPER
* BFS THEN DFS
* int overflow, array bounds
* special cases (n=1?)
* do sm th instead of nothing and stay organized
* DON'T GET STUCK ON ONE APPROACH
* (STUCK?)******** Try to simplify the problem(keeping in mind the main problem), ():
* 1- problem to subProblem
* 2- from simple to complex: start with a special
* problem and then try to update the solution for general case
* -(constraints - > solve it with none , one,two ... of them till you reach the given problem
-(no constraints - > try to give it some)
-how a special case may be incremented
* 3-Simplification by Assumptions
* REVERSE PROBLEM
* PROBLEM ABSTRACTION
* SMALL O BSERVATIONS MIGHT HELP ALOT
* WATCH OUT FOR TIME
* RETHINK YOUR IDEA,BETTER IDEA, APPROACH?
* CORRECT IDEA, NEED MORE OBSERVATIONS
* CORRECT APPROACH, WRONG IDEA
* WRONG APPROACH
* THINK CONCRETE THEN SYMBOL,
* having the solution for the first m state , can we solve it for m + 1 ?
* in many cases incremental thinking needs data sorting
*/
