/*
Baltic 2021 The short shank; Redemption
- Call prisoners with t[i] <= T "active" and others "passive"
- We basically want to maximize the number of passive prisoners
that don't protest
- Consider a forest where:
- The nodes are the passive prisoners
- The parent of node i is the first prisoner to the right of
i that won't rebel if there's a mattress to the left of i
- We can find this forest by sweeping through the prisoners and
using a stack
*/
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
int depth[2000001], mx_child[2000001];
vector<int> children[2000001];
int main() {
cin.tie(0)->sync_with_stdio(0);
int n, d, tx;
cin >> n >> d >> tx;
int ans = n, mx = 0;
priority_queue<pair<int, int>> pq;
stack<pair<int, int>> stck;
for (int i = 1; i <= n; i++) {
int t;
cin >> t;
if (t <= tx) mx = max(mx, i + tx - t);
else if (mx < i) ans--;
else {
depth[i] = 1;
while (stck.size()) {
if (stck.top().second > i)
break;
int node = stck.top().first;
children[i].push_back(node);
if (depth[node] + 1 > depth[i]) {
depth[i] = depth[node] + 1;
mx_child[node] = i;
}
stck.pop();
}
stck.push({i, mx});
}
}
while (stck.size()) {
int root = stck.top().first;
pq.push({depth[root], root});
stck.pop();
}
while (d-- && pq.size()) {
pair<int, int> curr = pq.top();
pq.pop();
ans -= curr.first;
int node = curr.second;
while (node) {
for (int i : children[node]) if (i != mx_child[node])
pq.push({depth[i], i});
node = mx_child[node];
}
}
cout << ans;
return 0;
}
