Codeforce417C——二分——Hamburgers

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 and n_c pieces of cheese in the kitchen. Besides, the shop nearby has 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 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 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 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 ≤ 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.

Sample Input

Input

BBBSSC
6 4 1
1 2 3
4

Output

2

Input

BBC
1 10 1
1 10 1
21

Output

7

Input

BSC
1 1 1
1 1 3
1000000000000

Output

200000000001

/*

题意  给你一种配方以及现在拥有的材料和每样材料的价钱和当前拥有的钱

问你最多能做出多少

二分  做出的数目。注意0x3f3f3f3f = 1e9 + 1e7 + ... 不超Int

*/

#include <cstdio> 

#include <cstring>

#include <algorithm>

using namespace std;



const int MAX = 110;

int numb, nums, numc;

char ss[MAX];

int b1, b2, s1, s2, c1, c2;

long long  cost, temp;



bool check(long long x)

{

    if(x * numb > b1){

      temp -=  b2 * (x * numb - b1);

    }

    if(x * nums > s1){

      temp -= s2 * (x *nums - s1);

    }

    if(x * numc > c1){

       temp -= c2 * (x * numc - c1);

    }

    //printf("%d\n", temp);

    if(temp < 0 ) return false;

    return true;

}

    

int main()

{

    while(~scanf("%s", ss)){

        scanf("%d%d%d", &b1, &s1, &c1);

        scanf("%d%d%d", &b2, &s2, &c2);

        scanf("%I64d", &cost);

        numb = nums = numc = 0;

        //printf("%I64d\n", cost);

        for(int i = 0 ; ss[i] != '\0'; i++){

            if(ss[i] == 'B') numb++;

            if(ss[i] == 'S') nums++;

            if(ss[i] == 'C') numc++;

        }

       // printf("%d %d %d\n", numb, nums, numc);

        long long l = 0, r = 1e14 + 5;

        long long ans = 0;

        while(l <= r){

            long long mid = (l + r) >> 1;

       //     printf("%lld\n", mid);

             temp = cost;

            if(check(mid)){

                l = mid + 1;

                ans = mid;

            }

            else r = mid - 1;

        }

        printf("%I64d\n", ans);

    }

    return 0;

}