Magic Powder - 2

D2. Magic Powder - 2

     time limit per test             1 second

     memory limit per test       256 megabytes

 

Waking up in the morning, Apollinaria decided to bake cookies. To bake one cookie, she needs n ingredients, and for each ingredient she knows the value ai — how many grams of this ingredient one needs to bake a cookie. To prepare one cookie Apollinaria needs to use all n ingredients.

Apollinaria has bi gram of the i-th ingredient. Also she has k grams of a magic powder. Each gram of magic powder can be turned to exactly 1 gram of any of the n ingredients and can be used for baking cookies.

Your task is to determine the maximum number of cookies, which Apollinaria is able to bake using the ingredients that she has and the magic powder.

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

 

题解 ： 二分法

 

 

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<long, long> PLL;
const ll mod = 1000000007;
const int N = 1e5+5; // 矩阵最大维数

ll n,m,k;
ll a[N],b[N],d[N];
ll ans; 

int f(ll d){
	ll kk=k;
	rep(i,0,n){
		if(a[i]*d>b[i]) kk-=(a[i]*d-b[i]);
		if( kk<0 ) return false;
	}
	return true;
}

int main() {
	cin>>n>>k;
	rep(i,0,n) cin>>a[i];
	rep(i,0,n) cin>>b[i];
	rep(i,0,n) ans+=a[i];
	ll l=0,r=2E9+1;
	ll res=0;ll mid;
	while(l<=r){
		mid = l+(r-l)/2;
		
		if(f(mid)){
			res=mid;
			l=mid+1;
		}
		else{
			r=mid-1;
		}
	}
	cout<<res<<endl;
}