#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define pb push_back
#define ff first
#define ss second
#define MOD 998244353LL
#define INF 1000000001LL
#define POT (1LL<<20)
#define INFL 1000000000000000099LL
#define pii pair<ll,ll>
#define ppi pair<pii,ll>
#define pip pair<ll,pii>
#define ppp pair<pii,pii>
#define vi vector<ll>
#define vii vector<pii>
#define al(x) x.begin(),x.end()
#define rev(x) reverse(al(x))
#define X 18
template<typename T, typename U>
pair<T, U> operator+(const pair<T, U>& a, const pair<T, U>& b) {
return {a.first + b.first, a.second + b.second};
}
template<typename T, typename U>
pair<T, U> operator-(const pair<T, U>& a, const pair<T, U>& b) {
return {a.first - b.first, a.second - b.second};
}
template<typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
os<<"{"<<p.ff<<", "<<p.ss<<"}";
return os;
}
template<typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
os << "{";
for (size_t i = 0; i < v.size(); ++i) {
if (i) os << ", ";
os << v[i];
}
os << "}";
return os;
}
ll fct[1000007];
ll fp(ll a,ll b){
ll c=1;while(b){if(b&1)c=(c*a)%MOD;a=(a*a)%MOD;b/=2;}
return c;
}
vi v,v2;
ll n,a,t,d;
vi g[2000007];
bool cz[2000007];
ll il[2000007];
ll dp[2000007];
ll dfs(ll v){
dp[v]=1;
for(ll i:g[v]){
dp[v]=max(dp[v],dfs(i)+1);
}
return dp[v];
}
void dfs2(ll v,bool b){
pii c={1,-1};
for(ll i : g[v])c=max(c,{dp[i]+1,i});
if(!b)il[c.ff]++;
for(ll i : g[v])dfs2(i,(i==c.ss));
}
void solve(){
cin>>n>>d>>t;
for(ll i=0;i<n;i++){
cin>>a;
v.pb(i+t-a);
}
ll mx=-1;
a=0;
for(ll i=0;i<n;i++){
mx=max(mx,v[i]);
v2.pb(mx);
if(mx>=i)cz[i]=1;
a+=cz[i];
}
stack<pii>s;
for(ll i=0;i<n;i++){
if(!cz[i])continue;
if(v[i]<i)
while(s.size() && s.top().ff<i){
if(v[s.top().ss]<s.top().ss)g[i].pb(s.top().ss);
s.pop();
}
s.push({v[i],i});
}
while(s.size()){
if(v[s.top().ss]<s.top().ss)
g[n].pb(s.top().ss);
s.pop();
}
dfs(n);
dfs2(n,0);
for(ll i=2000006;i>=0;i--){
while(d && il[i]){
d--;
a-=i;
il[i]--;
}
}
cout<<a+1;
}
int main(){
fct[0]=1;
for(ll i=1;i<1000007;i++)fct[i]=(fct[i-1]*i)%MOD;
ll t=1;
//cin>>t;
for(ll i=1;i<=t;i++){
//cout<<"Case #"<<i<<": ";
solve();
}
}
