Codeforces 371C Hamburgers

Description

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 n_b pieces of bread,n_s pieces of sausage andn_c pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices arep_b rubles for a piece of bread,p_s for a piece of sausage andp_c 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 EnglishB), 'S' (uppercase EnglishS) and 'C' (uppercase EnglishC).

The second line contains three integers n_b,n_s,n_c (1 ≤ n_b, n_s, n_c ≤ 100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integersp_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 integerr (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 thecin, cout streams or the%I64d specifier.

Output

Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print0.

Sample Input

Input

BBBSSC6 4 11 2 34

Output

2

Input

BBC1 10 11 10 121

Output

7

Input

BSC1 1 11 1 31000000000000

Output

200000000001

题意很好理解哒，思路也很清晰，可能实现起来略麻烦，分两步处理，先把已有的处理掉，再处理剩下的钱。

AC代码：

#include <cstdio>#include <iostream>#include <algorithm>#include <cmath>#include <cstring>#include <stdlib.h>using namespace std;typedef long long ll;char s[105];int nb,ns,nc;int pb,ps,pc;ll r;int main(){    scanf("%s",&s);    scanf("%d%d%d",&nb,&ns,&nc);    scanf("%d%d%d",&pb,&ps,&pc);    scanf("%I64d",&r);    int sb=0,ss=0,sc=0;    int len=strlen(s);    for(int i=0;i<len;i++){        if(s[i]=='B') sb++;        if(s[i]=='S') ss++;        if(s[i]=='C') sc++;    }    //printf("%d %d %d ",sb,ss,sc);    ll ans=0;    bool f1,f2,f3;    while(1){        f1=false,f2=false,f3=false;        if(nb-sb>=0) {f1=true;nb-=sb;}        if(ns-ss>=0) {f2=true;ns-=ss;}        if(nc-sc>=0) {f3=true;nc-=sc;}        //printf("%d %d %d\n",nb,ns,nc);        if(!f1&&(sb-nb)*pb<=r){            r-=(sb-nb)*pb;            f1=true;            nb=0;        }        if(!f2&&(ss-ns)*ps<=r){            r-=(ss-ns)*ps;            f2=true;            ns=0;        }        if(!f3&&(sc-nc)*pc<=r){            r-=(sc-nc)*pc;            f3=true;            nc=0;        }        if(f1&&f2&&f3) ans++;        if(!f1||!f2||!f3) break;        if(ns*ss==0&&nb*sb==0&&nc*sc==0) break;    }    if(r>0){        ll temp=r/(pb*sb+ps*ss+pc*sc);        ans+=temp;    }    printf("%I64d\n",ans);    return 0;}