#include <bits/stdc++.h>
#define div1 1000000007
const long long INF = 1LL << 60;
using namespace std;
typedef long long ll;
typedef long double ld;
// a>b -> a==b;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
// a<b -> a==b;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

int main(void)
{
    ll N, A, B, i, X[100001], ans=0;
    
    cin >> N >> A >> B;

    for(i=1; i<=N; i++){
        cin >> X[i];

        if(i>=2){
            ans += min(A*(X[i]-X[i-1]), B);
        }
    }

    cout << ans << endl;

    return 0;
}
#include<bits/stdc++.h>

using namespace std;

#define LL long long
#define DD double
#define Pb push_back
#define Bp pop_back
#define Pf push_front
#define Fp pop_front
#define Ub upper_bound
#define Lb lower_bound
#define In insert
#define All(x) x.begin(), x.end()
#define mem(a, b) memset(a, b, sizeof(a))
#define fast ios_base::sync_with_stdio(0);cin.tie(0)
#define X first
#define Y second
#define mx1 10
#define mx2 110
#define mx3 1010
#define mx4 10010
#define mx5 100010
#define mx6 1000010

typedef vector<LL> Vl;
typedef vector<DD> Vd;
typedef vector<bool> Vb;
typedef vector<Vl> VVl;
typedef pair<LL, LL> Pll;
typedef pair<DD, DD> Pdd;
typedef vector<Pll> Vpl;
typedef vector<Pdd> Vpd;
typedef queue<LL> Ql;
typedef stack<LL> Sl;
typedef deque<LL> Dl;

LL _set(LL N, LL pos) {return N = N | (1 << pos);}
LL _reset(LL N, LL pos) {return N = N & ~(1 << pos);}
bool _check(LL N, LL pos) {return (bool)(N & (1 << pos));}
bool _upper(char a) {return a >= 'A' && a <= 'Z';}
bool _lower(char a) {return a >= 'a' && a <= 'z';}
bool _digit(char a) {return a >= '0' && a <= '9';}

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

///**************************DONE****************************///

int main()
{
    LL n, t;
    cin >> n >> t;

    LL ans = 0, x = 0;
    for(int i = 0; i < n; i++){
        LL a;
        cin >> a;

        if(x < a)
            ans += t;
        else
            ans += a + t - x;
        x = a + t;
    }

    cout << ans << '\n';
    return 0;
}
