#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m; cin >> n >> m;
vector<int> a(n);
for (int &x: a) cin >> x;
/**
* Follows https://usaco.guide/problems/lmio-2019triusis/solution
*
* We count the number of poles we don't change
* Suppose the poles i1, i2, ..., ik don't change
* Condition 1. a[i_j] <= M * i_j
* Condition 2. a[i_{j+1}] <= a[i_j] + M * (i_{j+1} - i_j)
*
* Hence, we have a[i_{j+1}] - M * i_{j+1} <= a[i_j] - M * i_j
* or M * i_{j+1} - a[i_{j+1}] <= M * i_j - a[i_j]
*
* Now, if b[i] = M * i - a[i], we can find the largest non-decreasing
* sequence of b[i_j] for i_j in {i1, i2, ..., ik}
* where the the indices satisfy condition 1 (b[i] >= 0)
*/
vector<ll> lis;
for (ll i = 0, x; i < n; i++) {
x = 1LL * m * (i + 1) - a[i]; // b[i]
if (x < 0) continue;
if (lis.empty() or x >= lis.back()) lis.push_back(x);
else *lower_bound(lis.begin(), lis.end(), x + 1) = x;
}
cout << n - lis.size() << endl;
}
