#include <bits/stdc++.h>
using namespace std;

using pii = pair<int,int>;
using ll = long long;
#define rep(i, j) for(int i=0; i < (int)(j); i++)
#define repeat(i, j, k) for(int i = (j); i < (int)(k); i++)
#define all(v) v.begin(),v.end()
#define debug(x) cerr << #x << " : " << x << endl

template<class T> bool set_min(T &a, const T &b) { return a > b  ? a = b, true : false; }
template<class T> bool set_max(T &a, const T &b) { return a < b  ? a = b, true : false; }
// vector
template<class T> istream& operator >> (istream &is , vector<T> &v) { for(T &a : v) is >> a; return is; }
template<class T> ostream& operator << (ostream &os , const vector<T> &v) { for(const T &t : v) os << "\t" << t; return os << endl; }
// pair
template<class T, class U> ostream& operator << (ostream &os , const pair<T, U> &v) { return os << "<" << v.first << ", " << v.second << ">"; }

const int INF = 1 << 30;
const ll INFL = 1LL << 60;


class Solver {
  public:
    bool solve() {
        ll N, P; cin >> N >> P;
        vector<ll> H(N); cin >> H;
        auto gezan = [&] (int &i) {
            ll sum = 0;
            while(i + 1 < N and H[i] >= H[i + 1]) {
                sum += H[i] - H[i + 1];
                i++;                
            }
            return sum;
        };
        auto tozan = [&] (int &i) {
            ll sum = 0;
            while(i + 1 < N and H[i] <= H[i + 1]) {
                sum += H[i + 1] - H[i];
                i++;
            }
            return sum;
        };
        int i = 0;
        vector<pair<ll, ll>> arr;
        gezan(i);
        while(i + 1 < N) {
            arr.push_back(make_pair(tozan(i), arr.size()));
            arr.push_back(make_pair(gezan(i), arr.size()));
        }
        if(arr.size() > 0 and arr.back().first == 0) arr.pop_back();        
        sort(all(arr));
        vector<int> used(N);
        ll ans = 0;
        rep(i, arr.size()) {
            ll cost = arr[i].first;
            int idx = arr[i].second;
            if(used[idx]) continue;
            used[idx] = true;
            if(idx + 1 < N) used[idx + 1] = true;
            if(idx > 0) used[idx - 1] = true;
            ans += min(P * 2, cost);
        }
        cout << ans << endl;
        return 0;
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    Solver s;
    s.solve();
    return 0;
}