Magic Powder

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 100000000011000000000

Output

2000000000

Input

10 11000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 10000000001 1 1 1 1 1 1 1 1 1

Output

0

Input

3 12 1 411 3 16

Output

4

Input

4 34 3 5 611 12 14 20

Output

3

一个很标准的二分，但自己还是反应不过来，真的菜的抠脚啊，注意long long,不要超了范围。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;long long a[100001];long long b[100001];int main(){   long long n,k;   scanf("%I64d %I64d",&n,&k);   for(int i=1; i<=n; i++)   {       scanf("%I64d",&a[i]);   }   for(int i=1; i<=n; i++)   {       scanf("%I64d", &b[i]);   }  long long x = 0;  long long y = 2000000000;  long long minx = 0;  while(x <= y)  {      long long mid = (x+y)/2;      long long kk = k;      long long i;      for(i=1; i<=n; i++)      {        if(mid*a[i] > b[i])        kk -= (mid*a[i]-b[i]);        if(kk<0)break;      }      if(i==n+1) {x=mid+1;minx = max(minx,mid);}      else y=mid-1;  }  printf("%I64d\n",minx);    return 0;}

水波。