CodeForces 371C Hamburgers

C. Hamburgers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.

Polycarpus has nb pieces of bread, ns pieces of sausage and nc pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are pb rubles for a piece of bread, ps for a piece of sausage and pc for a piece of cheese.

Polycarpus has r rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient.

Input

The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C).

The second line contains three integers nb, ns, nc (1 ≤ nb, ns, nc ≤ 100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers pb, ps, pc (1 ≤ pb, ps, pc ≤ 100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer r (1 ≤ r ≤ 1012) — the number of rubles Polycarpus has.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Output

Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0.

Examples

input

BBBSSC6 4 11 2 34

output

2

input

BBC1 10 11 10 121

output

7

input

BSC1 1 11 1 31000000000000

output

200000000001

依旧是难啊难。本来想用模拟，结果差点没把我绕进去= =……最后还是用二分了。其实整数二分的判定还是不太清楚，参考了部分别人的题解……希望以后遇到类似的题能彻彻底底明白吧……

想出二分方法是关键。主要是把假设能做的需要花的价格和自己有的钱对比一下。

题意：坡旅甲很喜欢汉堡，他现在要做汉堡。第一行输入要做的汉堡的种类。B代表面包片，S代表肉饼，C代表奶酪。第二行是nb,ns,nc，分别表示他现在有的面包片，肉饼以及奶酪的数量。第三行pb,ps,pc，分别表示如果他去商店购入这些材料，分别需要多少钱。现在坡旅甲有r元钱，问，他最多能做多少汉堡。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int b,s,c,nb,ns,nc,pb,ps,pc;__int64 R;int OK(__int64 x){__int64 B,S,C;B=(x*b-nb)*pb;if(B<0)B=0;S=(x*s-ns)*ps;if(S<0)S=0;C=(x*c-nc)*pc;if(C<0)C=0;if(B+C+S>R)return 1;if(B+C+S==R) return -1;return 0;}char ch[111];int main(){int i;while(~scanf("%s",ch)){b=s=c=0;int L=strlen(ch);for(i=0;i<L;i++){if(ch[i]=='B')b++;else if(ch[i]=='S')s++;elsec++;}scanf("%d%d%d",&nb,&ns,&nc);scanf("%d%d%d",&pb,&ps,&pc);scanf("%I64d",&R);__int64 l=0,r=10000000000000;__int64 mid,ans;while(l<=r){mid=(l+r)/2;if(OK(mid)==-1){ans=mid;r=mid-1;}else if(OK(mid)==1){ans=mid-1;r=mid-1;}elsel=mid+1;}printf("%I64d\n",ans);}return 0;}