Codeforces Round #350 (Div. 2) D2 二分

D2. Magic Powder - 2

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

题意：http://www.cnblogs.com/hsd-/p/5464525.html

         （补）直通D1  数据范围变大

题解：二分处理

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll __int64
using namespace std;
ll n,k;
ll a[100005];
ll b[100005];
ll gg;
bool check(ll x)
{
    ll gg=k;
    for(int i=1;i<=n;i++)
    {
        if((x*a[i])>b[i])
        {
            ll cha=x*a[i]-b[i];
            gg=gg-cha;
        }
        if(gg<0)
            return false;
    }
    return true;
}
int main ()
{
    scanf("%I64d %I64d",&n,&k);
    ll gg=0;
    for(int i=1;i<=n;i++)
        scanf("%I64d",&a[i]);
    for(int i=1;i<=n;i++)
        scanf("%I64d",&b[i]);
     ll l=0,r=2000000000,mid;
     while(l<r)
     {
         //cout<<l<<" "<<r<<endl;
         mid=(l+r+1)>>1;
         if(check(mid))
            l=mid;
         else
            r=mid-1;
     }
     cout<<l<<endl;
    return 0;
}