https://atcoder.jp/contests/abc052/tasks/arc067_b
// https://atcoder.jp/contests/abc052/tasks/arc067_b
// 贪心即可, 从左到右行动, 每步选择代价小的方式
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 1e5 + 10;
LL n, a, b;
LL x[N];
void solv()
{
cin >> n >> a >> b;
LL ans = 0;
cin >> x[1];
for (int i = 2; i <= n; i ++)
{
cin >> x[i];
ans += min((x[i] - x[i-1]) * a, b);
}
cout << ans << endl;
}
int main()
{
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T --)
{
solv();
}
return 0;
}