题干：

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 ningredients 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.

 

解题报告：

      单单从这一道题上来看，先找出不用magic powder所能做的最大蛋糕数，然后模拟多做一个蛋糕，两个蛋糕....一直到magic powder小于等于0为止。但是因为这道题还有后续（）所以最好是用枚举去理解，也就是先找出不用magic powder所能做的最大蛋糕数，然后枚举做一个蛋糕，做两个蛋糕，，一直到magic powder小于等于0为止。然后 对于那个后续的题目，就是针对这一枚举的过程用二分优化，来加快查找速度。

AC代码：

#include<bits/stdc++.h>

using namespace std;
const int INF = 0x3f3f3f3f;
int a[1005],b[1005];
int main()
{
	int n,qq;
	cin>>n>>qq;
	for(int i = 1; i<=n; i++) scanf("%d",&a[i]);
	for(int i = 1; i<=n; i++) scanf("%d",&b[i]);
	int minn = INF;
	for(int i = 1; i<=n; i++) {
		minn = min(minn,b[i]/a[i]);
	}
	for(int i =1; i<=n; i++) {
		b[i]-=minn*a[i];
	}
	int ans = minn;
	int flag = 1;
	while(flag) {
		for(int i = 1; i<=n; i++) {
			if(b[i] < a[i]) {
				qq-=(a[i]-b[i]);
				if(qq >= 0) b[i] =a[i];
			}
			if(qq <= 0) {
				flag = 0;break;
			}
		}
		for(int i = 1; i<=n; i++) {
			if(b[i] < a[i]) {
				ans--;break;
			}
			b[i]-=a[i];
		}
		ans++;
	}
	printf("%d\n",ans);
	return 0 ;
}