// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



#define int long long

int n, a, b;

int x[101010];



int dp[101010];

int dfs(int i=1, int c=0)

{

        if (i == n){

                return 0;

        }



        int &ret = dp[i];

        if (ret != -1) return ret;

        ret = min(dfs(i + 1, c) + a * (x[i] - x[i - 1]), dfs(i + 1, c + 1) + b);

        return ret;

}

main()

{

        cin >> n >> a >> b;



        memset(dp, -1, sizeof(dp));

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

                cin >> x[i];

        }



        /*

        for (int i = 1; i < n; i++){

                ans += min(b, a * (x[i] - x[i - 1]));

        }

        */



        cout << dfs() << endl;

}




// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;



// clang-format off

#define range(i, l, r) for (int i = (int)l; i < (int)(r); ++(i))

#define rrange(i, l, r) for (int i = ((int)(r) - 1); i >= (int)l; --(i))



#define whole(f,x,...) ([&](decltype((x)) whole) { return (f)(begin(whole), end(whole), ## __VA_ARGS__); })(x)

#define rwhole(f,x,...) ([&](decltype((x)) whole) { return (f)(rbegin(whole), rend(whole), ## __VA_ARGS__); })(x)



#define debug(x) cerr << "(" << __LINE__ << ")" << #x << ": " << (x) << endl



inline void br(ostream &os = cerr) { os << endl; }



using int32 = int;

using int64 = long long;



const int mod     = 1e9 + 7;

const int32 inf   = 1001001001;

const int64 infll = 1001001001001001001ll;



const int dx[] = {0, -1, 1, 0, -1, 1, -1, 1}, dy[] = {-1, 0, 0, 1, -1, -1, 1, 1};

const string YES = "YES", NO = "NO", Yes = "Yes", No = "No", yes = "yes", no = "no";



struct IoSetup { IoSetup(){ cin.tie(0); ios::sync_with_stdio(0); cout << fixed << setprecision(10); cerr << fixed << setprecision(10); } } iosetup;



template <typename T = int64> T input() { T x; cin >> x; return x; }



template <typename T> ostream &operator<<(ostream &os, vector<T> &v) { range(i, 0, v.size()) { os << v[i] << (i + 1 != v.size() ? " " : ""); } return os; } 

template <typename T> istream &operator>>(istream &is, vector<T> &v) { for (T &in : v) is >> in; return is; }

template <typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p) { os << p.fs << " " << p.sc; return os; }

template <typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { is >> p.fs >> p.sc; return is; }



template <typename T> vector<T> make_v(size_t a, T b) { return vector<T>(a, b); }

template <typename... Ts> auto make_v(size_t a, Ts... ts) { return vector<decltype(make_v(ts...))>(a, make_v(ts...)); }



template <typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }

template <typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }

// clang-format on



void solver() {

  int64 N = input(), A = input(), B = input();



  vector<int64> X(N);

  cin >> X;



  int64 ans = 0;

  range(i, 1, N) {

    ans += min((X[i] - X[i - 1]) * A, B);

  }



  cout << ans << endl;

}



signed main(int argc, char *argv[]) {

  solver();

}

// Fastest known solution for this problem: