kuangbin——Hamburgers(二分答案)

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 %I64d specifier.

Output

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

Examples
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

二分答案解題思路：
二分答案的題目首先一定要滿足答案單調的性質，而本題答案製作漢堡的數量剛好具有這個性質於是考慮二分答案。
由題目分析可以得出上面的花費方程，進而以花費作為判斷條件二分答案。（這類題目和重要的一點在於區間劃分）

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
using ll = long long;

string hmb;
ll hb, hs, hc, nb, ns, nc, pb, ps, pc;
ll num = 0, cost = 0, money;

bool check(ll x)
{
	ll sum = max((ll)0, hb * x - nb) * pb + max((ll)0, hc * x - nc) * pc + max((ll)0, hs * x - ns) * ps;
	return sum <= money;
}

int main()
{
	cin >> hmb;
	cin >> nb >> ns >> nc >> pb >> ps >> pc >> money; // n為擁有的， h為需要的

	for (int i = 0; i < hmb.size(); i++)
	{
		if (hmb[i] == 'B') hb++;
		else if (hmb[i] == 'S') hs++;
		else hc++;
	}

	//二分查詢當前的料可以做幾個漢堡
	ll right = 1e13, left = 0, mid, ans = 0;
	while (left <= right) //區間存在問題
	{
		mid = right - (right - left) / 2;
		if (check(mid)) //如果可以做
		{
			ans = mid;
			left = mid + 1;
		}
		else right = mid - 1;
	}
	cout << ans;
	return 0;
}