Problem:
Score : 500 points

Problem Statement

There are N towns on a line running east-west.
The towns are numbered 1 through N, in order from west to east.
Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value.
The coordinate of town i is X_i.

You are now at town 1, and you want to visit all the other towns.
You have two ways to travel:

Walk on the line.
Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.

Teleport to any location of your choice.
Your fatigue level increases by B, regardless of the distance covered.

Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.

Constraints

All input values are integers.

2≤N≤10^5

1≤X_i≤10^9

For all i(1≤i≤N-1), X_i<X_{i+1}.

1≤A≤10^9

1≤B≤10^9

Input

The input is given from Standard Input in the following format:

N A B
X_1 X_2 ... X_N

Output

Print the minimum possible total increase of your fatigue level when you visit all the towns.

Sample Input 1

4 2 5
1 2 5 7

Sample Output 1

11

From town 1, walk a distance of 1 to town 2, then teleport to town 3, then walk a distance of 2 to town 4.
The total increase of your fatigue level in this case is 2×1+5+2×2=11, which is the minimum possible value.

Sample Input 2

7 1 100
40 43 45 105 108 115 124

Sample Output 2

84

From town 1, walk all the way to town 7.
The total increase of your fatigue level in this case is 84, which is the minimum possible value.

Sample Input 3

7 1 2
24 35 40 68 72 99 103

Sample Output 3

12

Visit all the towns in any order by teleporting six times.
The total increase of your fatigue level in this case is 12, which is the minimum possible value.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
 
// #define DEBUG
#define fi first
#define se second
#define be begin
#define en end
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define ALL(a) (a).be() , (a).en()
#define REP(i,n) for(int (i)=0;(i)<(n);(i)++)        //repeat n times
#define REP2(i,s,n) for(int (i)=(s);(i)<(n);(i)++)   //repeat from s to n 
#define REPD(i,n) for(int (i)=(n);(i)>=0;(i)--)      //repeat from n to 0
#define REPD2(i,s,e) for(int (i)=(s);(i)>=(e);(i)--) //repeat from s to e
#define RANGE(i,v) for(auto &(i):v)                  //repeat range
#define ASIZE(a) (sizeof(a) / sizeof(a[0]))          //array size
 
#define EL cout<<endl
 
using LL = long long;
 
template<typename T1,typename T2> using P = pair< T1, T2 >;
using Pii = P<int,int>;
using Pdd = P<double,double>;
 
template<typename T> using V = vector< T >;
using Vi = V<int>;
using Vll = V<LL>;
using Vs = V<string>;
 
template<typename T1,typename T2> using M = map< T1, T2>;
using Mii = M<int, int>;
using Msi = M<string, int>;
 
const int MOD = 1000000007;
const int INF = 1999999999;
const LL INFLL = 999999999999999LL;
const double EPS = 1e-10;
 
const int DX[8]= {-1,0,1,0,-1,-1,1,1};
const int DY[8]= {0,-1,0,1,-1,1,1,-1};
 
const double PI=3.141592653589793;
//==================================================================
/* I/O */
//scan
void SCAN(int *a){scanf("%d",a);}                             //scan for type int
void SCAN(int *a,int n){REP(i,n){scanf("%d",&a[i]);}}   //scan for type int array
void SCAN(Pii *a){scanf("%d",&a->fi);scanf("%d",&a->se);}     //scan for type pair<int, int>
void SCAN(LL *a){scanf("%lld",a);}                            //scan for type long long int
void SCAN(LL *a,int n){REP(i,n){scanf("%lld",&a[i]);}}  //scan for type long long int array
void SCAN(char *c){scanf(" %c",c);}                           //scan for type char
void SCAN(char *c,int n){REP(i,n){scanf(" %c",&c[i]);}} //scan for type char array
 
//print 
void PRINT(int a){printf("%d\n",a);}
void PRINT(int *a,int s,char c='\n'){REP(i,s){if(i==s-1){c='\n';}printf("%d%c",a[i],c);}}
void PRINT(Vi a,char c='\n'){REP(i,a.size()){if(i==a.size()-1){c='\n';}printf("%d%c",a[i],c);}}
void PRINT(LL a){printf("%lld\n",a);}
void PRINT(LL *a,int s,char c='\n'){REP(i,s){if(i==s-1){c='\n';}printf("%lld%c",a[i],c);}}
void PRINT(double a){printf("%.15f\n",a);}
void PRINT(double *a,int s,char c='\n'){REP(i,s){if(i==s-1){c='\n';}printf("%f%c",a[i],c);}}
void PRINT(char a){printf("%c\n",a);}
void PRINT(string a){printf("%s\n",a.c_str());}
//==================================================================
 
//==================================================================
/* vector */
//sort and remove duplicates for type vector
template<typename A>
void UNIQUE(vector< A > &a,int mode=0){
	if(mode==0){ sort(ALL(a),greater< A >());}else{ sort(ALL(a));}
	a.erase(unique(ALL(a)),a.en());
}
 
//fill an N-dimensional array with val
template<typename A, size_t N, typename T>
void FILL(A (&array)[N], const T &val){fill((T*)array,(T*)(array+N),val);}
//==================================================================
 
//==================================================================
/* math */
// sign of val
template <typename T> int sgn(T val){return (T(0) < val) - (val < T(0));}
 
long double pascalTri(int n,int r){
	long double tri[n+1][n+1];
	REP(i,n+1){REP(j,n+1){tri[i][j]=0;}}
	REP(i,n+1){REP(j,n+1){
		if(j>i){break;}
		if(j==0||j==i){tri[i][j]=1;}else{tri[i][j]=(tri[i-1][j-1]+tri[i-1][j]);}
	}}
	return tri[n][r];
}
 
LL GCD(LL a,LL b){
	LL t; LL r;
	if(a<b){t=a;a=b;b=t;}
	if(b==0){return a;}
	while(a%b!=0){r=a%b;a=b;b=r;}
	return b;
}
 
LL LCM(LL a,LL b){LL ab=(a*b)%MOD;return ab/GCD(a%b,b);}
 
//Bipartite method : x^n
LL BMPow(int x,int n,int m=0){
	LL ans=1;LL p=x;
	if(m==0){ while(n>0){if(n&1==1){ans*=p;} p*=p; n>>=1;}}
	else{ while(n>0){if(n&1==1){ans=(ans*p)%m;} p=(p*p)%m ;n>>=1;}}
	return ans;
}
 
//x^-1=x^(m-2) mod m
LL modInv(LL x,int m){return BMPow(x,m-2,m);}
 
//x! or x! mod m
LL factorial(int x,int m=0){
	LL a=1;
	if(m==0){REPD2(i,x,1){a*=i;}}else{REPD2(i,x,1){a=(a*i)%m;}}
	return a;
}
 
P<Vll,Vll> primeFactor(LL n){
	Vll p,e; LL m=n; int c;
	for(LL i=2;i*i<=n;i++){
		if(m%i!=0){continue;}
		c=0;
		while(m%i==0){c++;m/=i;}
		p.pb(i); e.pb(c);
	}
	if(m>1){ p.pb(m); e.pb(1);}
	return mp(p,e);
}
 
 
//==================================================================
 
//==================================================================
/* geometry */
template<typename T> using coordinate = P< T, T >;
template<typename T> using coordinateSet = V<coordinate< T >>;
 
//centroid of polygon
template<typename T>
coordinate<double> centroidPolygon(coordinateSet<T> &a){
	coordinate<double> G; G.fi=0.; G.se=0.;
	int n = a.size();
	RANGE(i,a){G.fi+=i.fi; G.se+=i.se;}
	G.fi/=n;G.se/=n;
	return G;
}
 
double area(coordinate<int> a,coordinate<int> b,coordinate<int> c){ 
	return ((b.fi-a.fi)*(c.se-a.se)-(b.se-a.se)*(c.fi-a.fi))/2.;
}
 
int checkCross(coordinate<int> v1,coordinate<int> v2,coordinate<int> a,coordinate<int> b){
	if(area(a,b,v1)*area(a,b,v2)>=0){return 0;}
	if(area(v1,v2,a)*area(v1,v2,b)>=0){return 0;}
	return 1;
}
//==================================================================
 
//==================================================================
/* graph */
struct edge{
	int src; int dst; int weight;
	edge() : src(0), dst(0), weight(0) {}
	edge(int s, int d, int w) : src(s), dst(d), weight(w) {}
};
 
using edges = V<edge>;
using graph = V<edges>;
 
void add_edge(edges &g,int s, int d, int w=1){g.pb(edge(s,d,w));}
void add_edge(graph &g,int s, int d, int w=1){g[s].pb(edge(s,d,w));}
 
V<Vi> floyd(const graph &g){
	int i,j,k; int n = g.size();
	V<Vi> dist(n,Vi(n,INF/2));
	REP(i,n){dist[i][i]=0;}
	REP(i,n){for(auto &e:g[i]){dist[e.src][e.dst]=e.weight;}}
	REP(k,n){REP(i,n){REP(j,n){dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);}}}
	return dist;
}
 
Vi dijkstra(graph &g,int s){
	int i,j,k; int n=g.size(); int visit[n]; Vi dist(n);
	priority_queue<Pii,V<Pii>,greater<Pii>> q;
	REP(i,n){visit[i]=0; dist[i]=INF/2;}
	dist[s]=0;
	q.push(mp(0,s));
	int nv; int min_cost=0;
	while(!q.empty()){
		int d,t; tie(d,t)=q.top(); q.pop();
		if(visit[t]==1){continue;}
		visit[t]=1; dist[t]=d;
		for(auto &e:g[t]){
			if(dist[e.dst]<=d+e.weight){continue;}
			q.push(mp(d+e.weight,e.dst));
		}
	}
	return dist;
}
//==================================================================
 
//==================================================================
/* data structure */
// binary indexed tree
template<typename T>
struct binaryIndexedTree{
	private:
		int n;
		V<T> x;
	public:
		binaryIndexedTree(int num=0) : n(num), x(n,0){}
		void add(int a, T w){for(int i=a;i<n;i|=i+1){ if(x[i]<w){x[i]=w;}}}
		T maximum(int a){ T m = -1;for(int k=a-1; k>=0;k=(k&(k+1))-1){ m=max(m,x[k]);} return m;}
};
 
// union find
struct unionFind{
	private:
		Vi data;
	public:
		unionFind(int size) : data(size,-1){}
		bool unionSet(int x,int y){
			x=root(x);y=root(y);
			if(x!=y){
				if(data[y]<data[x]){swap(x,y);}
				data[x]+=data[y]; data[y]=x;
			}
			return x!=y;
		}
		bool findSet(int x,int y){ return root(x)==root(y);}
		int root(int x){return ((data[x]<0)?x:(data[x]=root(data[x])));}
		int size(int x){return -data[root(x)];}
};
//==================================================================


int main()
{
	// cin.tie(0);
	// ios::sync_with_stdio(false);
	int i,j;
	int N,A,B;

	SCAN(&N); SCAN(&A); SCAN(&B);
	LL X[N];
	SCAN(X,N);

	LL cnt=0;
	REP2(i,1,N)
	{
		if((X[i]-X[i-1])*A<B)
		{
			cnt+=(X[i]-X[i-1])*A;
		}
		else
		{
			cnt+=B;
		}
	}

	PRINT(cnt);

	return 0;
}