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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var n,c:int64;
    i:longint;
    sum,ans,ans2:int64;
    firstl,firstr:longint;
    x,v,num:array[0..100500]of int64;
begin
  read(n,c);
  for i:=1 to n do
  begin
    read(x[i],v[i]);
    num[i]:=num[i-1]+v[i];
    inc(sum,v[i]);
    if sum-x[i]>ans then
    begin
      ans:=sum-x[i];
      firstl:=i;
    end;
  end;
  sum:=0;
  for i:=n downto 1 do
  begin
    inc(sum,v[i]);
    if sum-(c-x[i])>ans2 then
    begin
      ans2:=sum-(c-x[i]);
      //writeln(i);
    end;
  end;
  if ans2>ans then ans:=ans2;
  for i:=1 to n do
  if x[i]>=c div 2 then
  begin
    firstr:=i;
    firstl:=i-1;
    break;
  end;
  for i:=1 to n do
  if x[i]<c div 2 then
  begin
    while (x[firstr+1]-num[firstr]<x[firstr]-num[firstr-1])and(firstr<n)do inc(firstr);
    if num[i]+num[n]-num[firstr-1]-2*x[i]+x[firstr]-c>ans then
    ans:=num[i]+num[n]-num[firstr-1]-2*x[i]+x[firstr]-c;
  end;
  if firstl>0 then
  begin
  for i:=n downto 1 do
  if x[i]>=c div 2 then
  begin
    while (num[firstl-1]-x[firstl-1]<num[firstl]-x[firstl])and(firstl>1)do dec(firstl);
    if num[n]-num[i-1]+num[firstl]-2*(c-x[i])-x[firstl]>ans then
    ans:=num[n]-num[i-1]+num[firstl]-2*(c-x[i])-x[firstl];
  end;
  end;
  writeln(ans);
end.