Magic Powder

https://codeforces.com/contest/670/problem/D2

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

input

3 1
2 1 4
11 3 16

output

input

4 3
4 3 5 6
11 12 14 20

output

题意：同上一题，就是数开大了，得用二分 Hait: INF比较小，所以得用3000000000ll就行了

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
typedef  long long ll ;
ll need[100008];
ll have[100008];
ll tmpp[100008];
ll n,k;
ll Slove(ll mid)
{   ll sum=k;for(ll i=1;i<=n;i++){tmpp[i]=have[i]-mid*need[i];if(tmpp[i]<0)sum+=tmpp[i];if(sum<0)return 0;}return 1;
}
int main()
{while(~scanf("%I64d%I64d",&n,&k)){for(ll i=1;i<=n;i++){scanf("%I64d",&need[i]);}for(ll i=1;i<=n;i++){scanf("%I64d",&have[i]);}ll ans=0;ll mid;ll l=0;ll r=3000000000ll;while(r>=l){mid=(l+r)/2;if(Slove(mid)==1){ans=mid;l=mid+1;}else{r=mid-1;}}printf("%I64d\n",ans);}//cout << "Hello world!" << endl;return 0;
}

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