Magic Powder

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 needsn ingredients, and for each ingredient she knows the valuea_i — how many grams of this ingredient one needs to bake a cookie. To prepare one cookie Apollinaria needs to use alln ingredients.

Apollinaria has b_i gram of thei-th ingredient. Also she has k grams of a magic powder. Each gram of magic powder can be turned to exactly1 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, ..., a_n (1 ≤ a_i ≤ 1000), where the i-th number is equal to the number of grams of thei-th ingredient, needed to bake one cookie.

The third line contains the sequence b₁, b₂, ..., b_n (1 ≤ b_i ≤ 1000), where the i-th number is equal to the number of grams of thei-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.

Example

Input

3 12 1 411 3 16

Output

4

Input

4 34 3 5 611 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 index1 and 1 gram of magic powder to ingredient with the index3. 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：

要用n种面粉做蛋糕，有k点魔力，这k点魔力，1点魔力可以变成任意1种面粉

第二行给出做出一块蛋糕需要的每种面粉的数量

第三行给出每种面粉的数量

求最多可以做多少块蛋糕

代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node{int id;  //第id种面粉int val; //做成一块蛋糕需要的id面粉的数量int cnt;//最多可以制作的蛋糕数量}arr[1005];int num[1005];int cmp(node a,node b){return a.cnt<b.cnt;}int main(){int n,k;    scanf("%d%d",&n,&k);    for(int i=0;i<n;i++)    scanf("%d",&num[i]);    for(int i=0;i<n;i++){    int a;    scanf("%d",&a);    arr[i].id=i;    arr[i].val=a%num[i];    arr[i].cnt=a/num[i];    }    sort(arr,arr+n,cmp);    while(1){    int i;    for( i=0;i<n-1;i++)//寻找最少数量    if(arr[i].cnt<arr[i+1].cnt) break;    k-=num[arr[i].id]-arr[i].val;     arr[i].val=0;    if(k>=0) arr[i].cnt++;    if(k<=0) break;    }    int i;    for(i=0;i<n-1;i++)    if(arr[i].cnt<arr[i+1].cnt) break;    printf("%d\n",arr[i].cnt);return 0;}