Magic Powder - 2 CodeForces - 670D2 （二分）

 Magic Powder - 2

  CodeForces - 670D2 

The term of this problem is the same as the previous one, the only exception — increased restrictions.

Input

The first line contains two positive integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 10⁹) — 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 ≤ 10⁹), 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, ..., b_n (1 ≤ b_i ≤ 10⁹), 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.

Example

Input

1 1000000000
1
1000000000

Output

2000000000

Input

10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1

Output

0

Input

3 1
2 1 4
11 3 16

Output

4

Input

4 3
4 3 5 6
11 12 14 20

Output

3

code:

#include 
#include 
#include 
using namespace std;
typedef long long ll;
ll n,k,a[100002],b[100002];

int main(){
    cin >> n >> k;
    ll i;
    for(i = 0; i < n; i++){
        cin >> a[i];
    }
    for(i = 0; i < n; i++){
        cin >> b[i];
    }//输入
    ll left = 0,right = 2000000000;//使用二分，先定义一个最大边界，然后二分查找个数
    ll ans = 0;
    while(left <= right){
        ll mid = (left+right)/2;
        ll m = k;
        int flag = 1;
        for(i = 0;i < n; i++){//以此判断每种原料
            if(a[i] * mid > b[i]){//如果所需原料乘以个数大于所有的原料，使用万能原料
                if(m - (a[i] * mid - b[i]) >= 0){//如果万能原料大于等于所需，则求出剩余的万能原料
                    m -= (a[i] * mid - b[i]);
                }
                else{
                    flag = 0;
                    break;//否则说明这个个数是不可能做出来的，太多
                }
            }
            //else如果所需原料都小于所有的那就不用管了
        }
        if(flag){//说明这个个数可以，看看更大可不可以
           ans = mid;
           left = mid + 1;
        }
        else{//否则找小一点的
            right = mid - 1;
        }
    }
    printf("%lld\n",ans);
    return 0;
}