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: Runtime Error)

// --------------------<optimizations>--------------------
#pragma GCC optimize("O3")
//(UNCOMMENT WHEN HAVING LOTS OF RECURSIONS)\
#pragma comment(linker, "/stack:200000000")
//(UNCOMMENT WHEN TRYING TO BRUTEFORCE WITH A LOT OF LOOPS)\
#pragma GCC optimize("unroll-loops")

#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

#include<bits/stdc++.h>

using namespace std;

#define ll long long
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define pii pair<ll,ll>
#define vpii vector<pair<ll,ll> >
#define F first
#define S second
#define ld long double
#define built __builtin_popcountll
#define mst(a,i) memset(a,i,sizeof(a))
#define all(x) x.begin(),x.end()
#define itit(it,a) for(auto it=(a).begin(); it!=(a).end(); it++)
#define rep(i,a,b) for(ll i=a; i<b; i++)
#define repr(i,a,b) for(ll i=a; i>b; i--)
#define reprr(i,a,b) for(ll i=a; i>=b; i--)
#define pi 3.14159265358979323846264338327950288419716939937510582097494459230
ll max3(ll x,ll y,ll z) {return max(max(x,y),z);}
ll min3(ll x,ll y,ll z) {return min(min(x,y),z);}
const ll M=2e5+10,M2=1e6+10,mod=1e9+7,inf=1e17+10;
const int INF=1e9+7;
void add(int& a, int b) {a+=b;if(a>=mod){a-=mod;}}
#define trace1(x)			cerr<<#x<<": "<<x<<endl
#define trace2(x, y)			 cerr<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define trace3(x, y, z)			 cerr<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define trace4(a, b, c, d)		 cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl
#define trace5(a, b, c, d, e)		 cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<endl
#define trace6(a, b, c, d, e, f)	 cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<" | "<<#f<<": "<<f<<endl
int X[] = {0, 1, 0, -1};
int Y[] = {-1, 0, 1, 0};
ll power(ll x,ll n){ll result=1;while(n>0){if(n%2==1)result=(result*x)%mod; x=((x%mod)*(x%mod))%mod;n=n/2;}return result;}

ll n,a,b;
ll x[M];

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
    cin >> n >> a >> b;
	for(ll i=0; i<n; i++) cin >> x[i];
	ll ans=0;
	for(ll i=1; i<n; i++){
        ans+=min(b,a*(x[i]-x[i-1]));
	}
	cout << ans;
	return 0;
}
/* The judge is never wrong! Your code is buggy
   Look for:
 * * Read the problem carefully.
 * * Don't forget to sort(), mod, ll!!!!
 * * Initial value = +/- infinity instead of zero!!!
 * * an easier and alternate approach
 * * read the problem statement carefully
 * * if concept is correct and still WA, try with a different implementation
 * * special cases (n=1?)
 * * if you have no idea just guess the appropriate well-known algorithm instead of doing nothing :/
 */