Codeforces Round #456 (Div. 2) A. Tricky Alchemy

傳送門：http://codeforces.com/contest/912/problem/A

A. Tricky Alchemy

time limit per test1 second

memory limit per test256 megabytes

Problem Description

During the winter holidays, the demand for Christmas balls is exceptionally high. Since it’s already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.

Grisha needs to obtain some yellow, green and blue balls. It’s known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.

Right now there are A yellow and B blue crystals in Grisha’s disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.

Input

The first line features two integers A and B (0 ≤ A, B ≤ 109), denoting the number of yellow and blue crystals respectively at Grisha’s disposal.

The next line contains three integers x, y and z (0 ≤ x, y, z ≤ 109) — the respective amounts of yellow, green and blue balls to be obtained.

Output

Print a single integer — the minimum number of crystals that Grisha should acquire in addition.

Examples

input

4 3

2 1 1

output

2

input

3 9

1 1 3

output

1

input

12345678 87654321

43043751 1000000000 53798715

output

2147483648

Note

In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.

解題心得：

1. 很簡單的一個模拟題，題意也很簡單就不說了，在比賽的時候頭腦發昏，居然用算出的和加起來去減給出的晶塊數目的和，簡直想砍掉自己的腦袋。
2. 注意使用long long資料量有點大。

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

int main()
{
    ll a,b,x1,x2,x3;
    scanf("%lld%lld%lld%lld%lld",&a,&b,&x1,&x2,&x3);
    ll y1,y2;
    y1 = y2 = ;//記錄需要使用的晶塊的數量

    y1 += x1*;
    y2 += x2;
    y1 += x2;
    y2 += x3*;

    ll ans = ;
    if(y1 > a)
        ans += y1-a;
    if(y2 > b)
        ans += y2-b;
    printf("%lld",ans);
    return ;
}