#include <bits/stdc++.h>   using namespace std;   vector<long long int> fact(2000006);   vector<long long int> prime_check(1e7 + 2);   vector<long long int> primes;   long long powmod(long long x, long long y, long long m) {     long long res = 1LL;     while (y) {       if (y & 1) res = (res * x) % m;       x = (x * x) % m;       y /= 2;     }     return res;   }   void init_fact() {     fact[0] = 1;     for (long long int i = 1; i < 2000006; i++) {       fact[i] = (fact[i - 1] * i) % 1000000007;     }   }   long long int C(long long int n, long long int r) {     if (n < r) return 0;     long long int temp =         fact[n] * powmod(fact[r], 1000000007 - 2, 1000000007) % 1000000007;     temp *= powmod(fact[n - r], 1000000007 - 2, 1000000007);     temp %= 1000000007;     return temp;   }   void Sieve() {     for (long long int i = 2; i <= 1e7 + 1; i++) {       if (prime_check[i]) continue;       primes.push_back(i);       for (long long int j = 2 * i; j <= 1e7 + 1; j += i) prime_check[j] = i;     }   }   long long int phi(long long int n) {     long long int i, res = n;     for (i = 2; i * i <= n; i++)       if (n % i == 0) {         while (n % i == 0) n /= i;         res -= res / i;       }     if (n > 1) res -= res / n;     return res;   }   long long int n, x, y;   vector<long long int> a(2000006), b(2000006), cnt(2000006), sum(2000006);   int main() {     ios_base::sync_with_stdio(false);     cin.tie(NULL);     cout.tie(NULL);     cin >> n >> x >> y;     for (long long int i = 1; i <= n; i++) cin >> a[i], b[a[i]]++;     for (long long int i = 1; i < 2000006; i++)       cnt[i] = cnt[i - 1] + b[i], sum[i] = sum[i - 1] + b[i] * i;     long long int c = x / y;     long long int ans = INT64_MAX;     for (long long int i = 2; i <= 1e6; i++) {       long long int cost = 0, f = i - min(i, c);       for (long long int j = 0; j <= 1e6; j += i) {         long long int st = j + 1, en = j + i, st2 = max(st, st + f - 1);         long long int c1 = cnt[st2 - 1] - cnt[st - 1];         long long int c2 = cnt[en] - cnt[st2 - 1];         long long int c3 = sum[en] - sum[st2 - 1];         cost += c1 * x + y * (c2 * en - c3);       }       ans = min(ans, cost);     }     cout << ans <<   n ;   }   