#include <bits/stdc++.h>
#include <atcoder/all>
typedef long long int ll;
typedef long long int ull;
#define MP make_pair
using namespace std;
using namespace atcoder;
typedef pair<ll, ll> P;
const ll MOD = 998244353;
// const ll MOD = 1000000007;
using mint = modint998244353;
// using mint = modint1000000007;
const double pi = 3.1415926536;
const int MAX = 2000003;
long long fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
void comp(vector<ll>&a){
  set<ll>s(a.begin(),a.end());
  map<ll,ll>d;
  ll cnt=0;
  for(auto x:s)d[x]=cnt++;
  for(auto&x:a)x=d[x];
}
ll gcd (ll x, ll y) {
   if (y == 0) return x;
   else if (y > x) {
       return gcd (y, x); 
   }
   else return gcd(x % y, y);
}
ll lcm (ll x, ll y) {
   return x / gcd(x, y) * y;
}
ll my_sqrt(ll x) {
    ll m = 0;
    ll M = 1000000001;
    while (M - m > 1) {
        ll now = (M + m) / 2;
        if (now * now <= x) {
            m = now;
        }
        else {
            M = now;
        }
    }
    return m;
}
ll keta(ll n) {
    ll ret = 0;
    while (n) {
        n /= 10;
        ret++;
    }
    return ret;
}
ll ceil(ll n, ll m) {
    ll ret = n / m;
    if (n % m) ret++;
    return ret;
}
ll pow_ll(ll x, ll n) {
    if (n == 0) return 1;
    if (n % 2) {
        return pow_ll(x, n - 1) * x;
    }
    else {
        ll tmp = pow_ll(x, n / 2);
        return tmp * tmp;
    }
}

int main() {
    ll n;
    ll k, x, y;
    cin >> n >> k >> x >> y;
    ll a[n];
    for (ll i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    for (ll i = 0; i < n; i++) a[i] = ceil(a[i] - 1, k);
    ll sum[n + 1];
    sum[0] = 0;
    for (ll i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i - 1];
    ll ans = 1000000000000000000;
    for (ll i = 0; i < n; i++) {
        ll tmp = 0;
        tmp += y * a[i];
        tmp += x * (sum[n] - sum[i + 1] - a[i] * (n - i - 1));
        ans = min(ans, tmp);
    }
    cout << min(ans, sum[n] * x);
    return 0;
}