Problem:
Score : 300 points

Problem Statement

"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively.

Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.

Constraints

1 ≤ A, B, C ≤ 5000

1 ≤ X, Y ≤ 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C X Y

Output

Print the minimum amount of money required to prepare X A-pizzas and Y B-pizzas.

Sample Input 1

1500 2000 1600 3 2

Sample Output 1

7900

It is optimal to buy four AB-pizzas and rearrange them into two A-pizzas and two B-pizzas, then buy additional one A-pizza.

Sample Input 2

1500 2000 1900 3 2

Sample Output 2

8500

It is optimal to directly buy three A-pizzas and two B-pizzas.

Sample Input 3

1500 2000 500 90000 100000

Sample Output 3

100000000

It is optimal to buy 200000 AB-pizzas and rearrange them into 100000 A-pizzas and 100000 B-pizzas. We will have 10000 more A-pizzas than necessary, but that is fine.

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

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

#define int long long
typedef long long ll;
const int N = 3005;
int n, M, s[N][N], c[N][N];

int fpow(int a, int b, int p) {
	int res = 1;
	for (; b; b >>= 1, a = a * 1ll * a % p) if (b & 1) res = res * 1ll * a % p;
    return res;
}

signed main() {
    scanf("%lld %lld", &n, &M);
    for (int i = 0; i <= n; i++) {
        c[i][0] = c[i][i] = 1;
        for (int j = 1; j < i; j++) c[i][j] = ((ll)c[i - 1][j] + (ll)c[i - 1][j - 1]) % M;
    }
    s[0][0] = 1;
    for (int i = 1; i <= n + 1; i++)
    	for (int j = 1; j <= i; j++) s[i][j] = ((ll)s[i - 1][j - 1] + 1ll * j * s[i - 1][j] % M) % M;
    int ans = 0;
    for (int i = 0; i <= n; i++) {
        int res = c[n][i];
        if ((i % 2) != 0) res = (M - res) % M; 
        int sum = 0, y = 1, kind = fpow(2, n - i, M);
        for (int k = 0; k <= i; k++) {
            sum = (sum * 1ll + s[i + 1][k + 1] * 1ll * y % M) % M; 
            y = y * 1ll * kind % M;
        }
        sum = (ll)sum * fpow(2, fpow(2, n - i, M - 1) + M - 1, M);
        ans = ((ll)ans + res * 1ll * sum % M) % M;
    }
    printf("%lld\n", ans);
    return 0;
}