CF912A Tricky Alchemy-白红宇

CF912A Tricky Alchemy

阅读量：6453 次

发布时间：2019-06-23

本文共 3367 字，大约阅读时间需要 11 分钟。

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

Copy

4 3 2 1 1

Output

Copy

Input

Copy

3 9 1 1 3

Output

Copy

Input

Copy

12345678 87654321 43043751 1000000000 53798715

Output

Copy

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.

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                         //#pragma GCC optimize(2)using namespace std;#define maxn 300005#define inf 0x3f3f3f3f#define INF 9999999999#define rdint(x) scanf("%d",&x)#define rdllt(x) scanf("%lld",&x)#define rdult(x) scanf("%lu",&x)#define rdlf(x) scanf("%lf",&x)#define rdstr(x) scanf("%s",x)typedef long long ll;typedef unsigned long long ull;typedef unsigned int U;#define ms(x) memset((x),0,sizeof(x))const long long int mod = 1e9 + 7;#define Mod 1000000000#define sq(x) (x)*(x)#define eps 1e-3typedef pair
                         
                           pii;#define pi acos(-1.0)const int N = 1005;#define REP(i,n) for(int i=0;i<(n);i++)typedef pair
                          
                            pii;inline ll rd() { ll x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x;}ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b);}ll sqr(ll x) { return x * x; }/*ll ans;ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans;}*/ll qpow(ll a, ll b, ll c) { ll ans = 1; a = a % c; while (b) { if (b % 2)ans = ans * a%c; b /= 2; a = a * a%c; } return ans;}int main(){ //ios::sync_with_stdio(0); ll a, b, x, y, z; rdllt(a); rdllt(b); rdllt(x); rdllt(y); rdllt(z); ll w1 = 2 * x + y - a; ll w2 = y + 3 * z - b; if (w1 > 0 && w2 > 0)cout << w1 + w2 << endl; else if (w1 > 0 && w2 <= 0)cout << w1 << endl; else if (w2 > 0 && w1 <= 0)cout << w2 << endl; else cout << 0 << endl; return 0;}

转载于:https://www.cnblogs.com/zxyqzy/p/10123496.html

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