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

A. Tricky Alchemy

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

第一次打cf。。。。。。
分析：签到题。

#include<cstdio>
int main()
{
    long long A,B,x,y,z;
    scanf("%lld%lld%lld%lld%lld",&A,&B,&x,&y,&z);
    long long yellow=x*2+y,blue=y+3*z;
    long long ans1=yellow-A,ans2=blue-B,ans=0;
    if(ans1>0) ans+=ans1;
    if(ans2>0) ans+=ans2;
    printf("%lld
",ans);
    return 0;
}