// Original C++ solution (slower):
#include <iostream>

#include <vector>

#include <cstdio>

#include <sstream>

#include <map>

#include <string>

#include <algorithm>

#include <queue>

#include <cmath>

#include <functional>

#include <set>

#include <ctime>

#include <random>

#include <chrono>

#include <cassert>

#include <tuple>

#include <utility>

using namespace std;



namespace {

  using Integer = long long; //__int128;

  template<class T, class S> istream& operator >> (istream& is, pair<T,S>& p){return is >> p.first >> p.second;}

  template<class T> istream& operator >> (istream& is, vector<T>& vec){for(T& val: vec) is >> val; return is;}

  template<class T> istream& operator ,  (istream& is, T& val){ return is >> val;}

  template<class T, class S> ostream& operator << (ostream& os, const pair<T,S>& p){return os << p.first << " " << p.second;}

  template<class T> ostream& operator << (ostream& os, const vector<T>& vec){for(size_t i=0; i<vec.size(); i++) os << vec[i] << (i==vec.size()-1?"":" "); return os;}

  template<class T> ostream& operator ,  (ostream& os, const T& val){ return os << " " << val;}



  template<class H> void print(const H& head){ cout << head; }

  template<class H, class ... T> void print(const H& head, const T& ... tail){ cout << head << " "; print(tail...); }

  template<class ... T> void println(const T& ... values){ print(values...); cout << endl; }



  template<class H> void eprint(const H& head){ cerr << head; }

  template<class H, class ... T> void eprint(const H& head, const T& ... tail){ cerr << head << " "; eprint(tail...); }

  template<class ... T> void eprintln(const T& ... values){ eprint(values...); cerr << endl; }



  class range{ Integer start_, end_, step_; public: struct range_iterator{ Integer val, step_; range_iterator(Integer v, Integer step) : val(v), step_(step) {} Integer operator * (){return val;} void operator ++ (){val += step_;} bool operator != (range_iterator& x){return step_ > 0 ? val < x.val : val > x.val;} }; range(Integer len) : start_(0), end_(len), step_(1) {} range(Integer start, Integer end) : start_(start), end_(end), step_(1) {} range(Integer start, Integer end, Integer step) : start_(start), end_(end), step_(step) {} range_iterator begin(){ return range_iterator(start_, step_); } range_iterator   end(){ return range_iterator(  end_, step_); } };



  inline string operator "" _s (const char* str, size_t size){ return move(string(str)); }

  constexpr Integer my_pow(Integer x, Integer k, Integer z=1){return k==0 ? z : k==1 ? z*x : (k&1) ? my_pow(x*x,k>>1,z*x) : my_pow(x*x,k>>1,z);}

  constexpr Integer my_pow_mod(Integer x, Integer k, Integer M, Integer z=1){return k==0 ? z%M : k==1 ? z*x%M : (k&1) ? my_pow_mod(x*x%M,k>>1,M,z*x%M) : my_pow_mod(x*x%M,k>>1,M,z);}

  constexpr unsigned long long operator "" _ten (unsigned long long value){ return my_pow(10,value); }



  inline int k_bit(Integer x, int k){return (x>>k)&1;} //0-indexed



  mt19937 mt(chrono::duration_cast<chrono::nanoseconds>(chrono::steady_clock::now().time_since_epoch()).count());



  template<class T> string join(const vector<T>& v, const string& sep){ stringstream ss; for(size_t i=0; i<v.size(); i++){ if(i>0) ss << sep; ss << v[i]; } return ss.str(); }



  inline string operator * (string s, int k){ string ret; while(k){ if(k&1) ret += s; s += s; k >>= 1; } return ret; }

}

constexpr long long mod = 9_ten + 7;





int main(){

  int n;

  long long e,t;

  cin >> n,e,t;



  vector<long long> x(n);

  for(auto& xx : x){

    scanf("%lld", &xx);

  }

  x.push_back(e);

  x.push_back(1ll<<57);



  vector<long long> dp(n+1, 1ll<<57);

  dp[0] = 0;



  deque<int> t_min;

  t_min.push_back(0);

  long long w = 1ll<<57;

  long long add = 0;



  for(int i=0, j=0; i<n; i++){

    while(j<i && t < 2*(x[i] - x[j]) ){

      if(w + add > dp[j] + 2*(x[i]-x[j])){

        w = dp[j];

        add = 2*(x[i] - x[j]);

      }

      j++;

    }

    while(t_min.size() && t_min.front() < j){

      t_min.pop_front();

    }



    dp[i+1] = min(dp[i+1], w+add);

    if(t_min.size()) dp[i+1] = min(dp[i+1], dp[t_min.front()] + t);



    add += 2*(x[i+1] - x[i]);

    while(t_min.size() && dp[t_min.back()] >= dp[i+1]){

      t_min.pop_back();

    }

    t_min.push_back( i+1 );

  }

  println(dp[n] + e);



  //eprintln(join(dp, " "));



  return 0;

}


// Optimized solution to the same problem (faster):
#include <vector>

#include <cstdio>

using namespace std;



int n;

long long e,t;

long long x[100005];

long long dp[100005];



int main(){

  scanf("%d%lld%lld", &n,&e,&t);

  for(int i=0; i<n; ++i){

    scanf("%lld", &x[i]);

    x[i]<<=1;

  }

  x[n] = e<<1;



  long long w = 1ll<<57;

  for(int i=0, j=0; i<n; ++i){

    while(j<i && t < (x[i] - x[j]) ){

      w = min(w, dp[j] + (x[i] - x[j]) );

      j++;

    }

    dp[i+1] = min(w, dp[j] + max((x[i]-x[j]) ,t));

    w += (x[i+1] - x[i]);

  }

  printf("%lld\n", dp[n] + e);

  return 0;

}


// Fastest known solution for this problem: