Codefroces D2. Magic Powder - 2(二分）

http://codeforces.com/problemset/problem/670/D2

http://codeforces.com/problemset/problem/670/D1

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

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
二分模板

#include<iostream>
using namespace std;
int binary_search(int a[],int l,int r,int k)
{
    int mid,ans;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(a[mid]<=k){
            l=mid+1;
            ans=mid;
        }
        else r=mid-1;
    }
    return ans;//也可ans+n;
}
int main()
{
    int a[11]={1,2,4,7,7,7,7,8,10,11};
    cout<<binary_search(a,0,9,4)<<endl;
    return 0;
}

题解

#include<iostream>
#include<cstdio>
using namespace std;
long long a[100006],b[100006];
int main()
{
    long long n,k,i;
    scanf("%lld%lld",&n,&k);
    for(i=0;i<n;i++)
    {
        scanf("%lld",&a[i]);
    }
    for(i=0;i<n;i++)
    {
        scanf("%lld",&b[i]);
    }
    long long l=0,r=2000000000;
    while(l<=r)
    {
        long long mid=(l+r)>>1;
        long long sum=0;
        for(i=0;i<n;i++)
        {
            if(mid*a[i]>b[i]) sum+=(mid*a[i]-b[i]);
            if(sum>k) break;
        }
        if(sum<=k) l=mid+1;
        else r=mid-1;
    }
    printf("%lld",l-1);
    return 0;
}