Codeforces Round #350 (Div. 2) D2 - Magic Powder - 2

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, ..., an (1 ≤ ai ≤ 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, ..., bn (1 ≤ bi ≤ 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：

做蛋糕，给出n种原料每种需要多少个才能做一个蛋糕，再给出当前每种原料的库存，最后给你魔法值可以拿来当做任何一种原料，问你最多做几个。

分析：

数据比较大，二分一下就好了。本来水题没什么记录的必要性，这里写二分的时候写二逼了，mid写在前面导致二分0，1时出现bug。D1题里面数据比较弱没触发这个bug，D2里面的第138组数据触发了，debug半天没想到是二分写错了。引以为戒。

#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <fstream>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
#define INF 0x3f3f3f3f
const int N=100005;
const int mod=1e9+7;

int a[N],b[N],c[N],n,k;

long long solve(int t) {
    long long sum=0,temp;
    for (int i=1; i<=n; i++) {
        temp=(long long)(t*a[i]-b[i]);
        if (temp>0) {
            sum+=temp;
        }
    }
    return sum;
}

int main() {
    long long sum;
    while (cin>>n>>k) {
        sum=0;
        for (int i=1; i<=n; i++) {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        int minn=INF,maxx=0;
        for (int i=1; i<=n; i++) {
            scanf("%d",&b[i]);
            c[i]=b[i]/a[i];
            minn=min(c[i],minn);
            maxx=max(maxx,c[i]);
        }
        int left=minn,right=maxx,mid=left+(right-left)/2,flag = minn+1;
        long long temp;
        temp=solve(maxx+1);
        if (temp<=k) {
            right=minn;
            flag=maxx+1;
        }
        while (left<right) {
            temp=solve(mid);
            if (temp==k) {
                break;
            } else if (temp>k) right=mid;
            else left=mid+1;
            mid=left+(right-left)/2;
            flag=mid;
        }
        if (solve(flag)>k) {
            flag--;
        }
        k-=solve(flag);
        int s=flag;
        if (k>sum) {
            s+=k/sum;
        }
        cout<<s<<endl;
    }
    return 0;
}