Hamburgers assuming the solution is feasible

HamburgersTime limit:1000MS Memory Limit:262144KB 64bit IO Format:%i64d &%i6 4u SubmitStatus

Description

Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there is 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" represent s the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.

Polycarpus has  n _{b  pieces of bread,   n s  pieces of sausage And  n c  pieces of cheese in the kitchen. Besides, the shop nearby have all three ingredients, the prices Are  p b  rubles for a piece of Bread,  p s  for a piece of sausage and  P c  for a piece of cheese.}

Polycarpus has r rubles and he are ready-to-shop on them. What maximum number of hamburgers can he cook? You can assume this polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has a 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, the string contains only letters 'B ' (uppercase 中文版b), 'S ' ( Uppercase 中文版S) and 'C ' (uppercase English C).

The second line contains three integersN_{b,Ns,NC(1≤Nb,N s , n C ≤100)-the number of the pieces of B Read, sausage and cheese on polycarpus ' kitchen. The third line contains three Integers  p b ,   p s ,   P C   (1≤ p b , p s , p C ≤100)- The price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains Integer  R   (1≤ R ≤10^{12)-the number of rubles Polycarpus has.}}

Please, don't write the %lld specifier to read or write 64-bit integers inс++. 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.

Sample Input

Input

Bbbssc
6 4 1
1 2 3
4

Output

2

Input

Bbc
1 10 1
1 10 1
21st

Output

7

Input

Bsc
1 1 1
1 1 3
1000000000000

Output

200000000001

1#include <stdio.h>2#include <string.h>3 4 Const intinf=0x3f3f3f3f;5 Const Long Longinf=1e14+7;6 Chara[ the];7 Long LongB,s,c,nb,ns,nc,pb,ps,pc,r;8 9 BOOLCLong Longx)Ten { One     Long LongRB,RS,RC; Arb= (X*B-NB) >=0? (X*B-NB):0, rs= (X*s-ns) >=0? (X*s-ns):0, rc= (X*C-NC) >=0? (X*C-NC):0; -     if((RB*PB+RS*PS+RC*PC) <=R) -         return true; the     Else -         return false; - } -  + intMain () - { +     inti,j,k; A      while(SCANF ("%s", a)! =EOF) at     { -b=0, s=0, c=0; -          for(i=0; a[i]!=' /'; i++) -         { -             if(a[i]=='B') -b++; in             Else if(a[i]=='S') -s++; to             Else if(a[i]=='C') +C++; -         } the         //printf ("%d%d%d", b,s,c); *scanf"%i64d%i64d%i64d%i64d%i64d%i64d",&nb,&ns,&nc,&pb,&ps,&pc); $scanf"%i64d",&R);Panax Notoginseng         Long Longlb=0, ub=INF; -  the          while(ub-lb>1) +         { A             Long LongMid= (LB+UB)/2; the             if(C (mid)) +lb=mid; -             Else $ub=mid; $         } -  -printf"%i64d\n", LB); the     } - Wuyi     return 0; the}

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Hamburgers assuming the solution is feasible