Codeforces Round #350 (Div. 2)（D）优先队列

D1. Magic Powder - 1

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraints, then you can just write a single solution to the both versions. If you find the problem too difficult in large constraints, you can write solution to the simplified version only.

Waking up in the morning, Apollinaria decided to bake cookies. To bake one cookie, she needs n ingredients, and for each ingredient she knows the value ai — how many grams of this ingredient one needs to bake a cookie. To prepare one cookie Apollinaria needs to use all n ingredients.

Apollinaria has bi gram of the i-th ingredient. Also she has k grams of a magic powder. Each gram of magic powder can be turned to exactly 1 gram of any of the n ingredients and can be used for baking cookies.

Your task is to determine the maximum number of cookies, which Apollinaria is able to bake using the ingredients that she has and the magic powder.

Input

The first line of the input contains two positive integers n and k (1 ≤ n, k ≤ 1000) — the number of ingredients and the number of grams of the magic powder.

The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000), where the i-th number is equal to the number of grams of the i-th ingredient, needed to bake one cookie.

The third line contains the sequence b1, b2, ..., bn (1 ≤ bi ≤ 1000), where the i-th number is equal to the number of grams of the i-th ingredient, which Apollinaria has.

Output

Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder.

Examples

input

3 1
2 1 4
11 3 16

output

4

input

4 3
4 3 5 6
11 12 14 20

output

3

Note

In the first sample it is profitably for Apollinaria to make the existing 1 gram of her magic powder to ingredient with the index 2, then Apollinaria will be able to bake 4 cookies.

In the second sample Apollinaria should turn 1 gram of magic powder to ingredient with the index 1 and 1 gram of magic powder to ingredient with the index 3. Then Apollinaria will be able to bake 3 cookies. The remaining 1 gram of the magic powder can be left, because it can't be used to increase the answer.

题意：要做饼干，每个饼干需要m种不同的的原料，每种原料需要a[i],你现在每种原料有b[i]，你现在还有K个魔法粉，一个魔法粉可以转化为任何一个单位的原料，问最多做多少饼干？

题解：显然，每次优先选择使用b[i]/a[i]最小的那个加上魔法粉，使用优先队列维护，然后找出最小的b[i]/a[i]输出

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cmath>  
#include<iostream>  
#include<algorithm>  
#include<vector>  
#include<map>  
#include<set>  
#include<queue>  
#include<string>  
#include<bitset>  
#include<utility>  
#include<functional>  
#include<iomanip>  
#include<sstream>  
#include<ctime>  
using namespace std;

#define inf int(0x3f3f3f3f)  
#define mod int(1e9+7)  
typedef long long LL;
enum{ N = int(2e5 + 10) };
int a[N], b[N], c[N];
struct point
{
	int x, y;
	int num;
	bool operator<(const point& bb)const
	{
		return num > bb.num;
	}
	point(int _x, int _y, int _num) :x(_x), y(_y), num(_num){}
	point(){}
};

priority_queue<point>q;
int main()
{
#ifdef CDZSC  
	freopen("i.txt", "r", stdin);
	//freopen("o.txt", "w", stdout);
	int _time_jc = clock();
#endif  
	int n, k;
	while (~scanf("%d%d", &n, &k))
	{
		while (!q.empty())q.pop();
		for (int i = 0; i < n; i++)
		{
			scanf("%d", a + i);
		}
		for (int i = 0; i < n; i++)
		{
			scanf("%d", b + i);
		}
		for (int i = 0; i<n; i++)
		{
			q.push(point(a[i], b[i], b[i] / a[i]));
		}
		point top;
		while (k>0)
		{
			top = q.top(); q.pop();
			top.y += 1;
			k--;
			top.num = top.y / top.x;
			q.push(top);
		}
		int ans = inf;
		while (!q.empty())
		{
			ans = min(ans, q.top().num);
			q.pop();
		}
		printf("%d\n", ans);
	}
	return 0;
}