Magic Powder - 2 （CF 670_D）

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

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

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
using namespace std;

#define N 110000
#define met(a, b) memset(a, b, sizeof(a))
#define INF 0x3f3f3f3f
const long long  Max = 2000000000;
typedef long long LL;


LL a[N], b[N];
LL n, k;

LL Judge(LL mid)
{
    LL i, K1=k, K2=k;

    for(i=1; i<=n; i++)
    {
        if(b[i]<a[i]*mid)
        {
            K1 -= (a[i]*mid - b[i]);
            if(K1<0)
                return -1;
        }
    }

    for(i=1; i<=n; i++)
    {
        if(b[i]<a[i]*(mid+1))
        {
            K2 -= (a[i]*(mid+1) - b[i]);
            if(K2<0)
                return 0;
        }
    }

    return 1;
}

int main()
{


    while(scanf("%I64d%I64d", &n, &k)!=EOF)
    {
        LL i;
        LL mid, L=0, R=Max, ans;

        met(a, 0);
        met(b, 0);

        for(i=1; i<=n; i++)
            scanf("%I64d", &a[i]);
        for(i=1; i<=n; i++)
            scanf("%I64d", &b[i]);

        while(L<R)
        {
            mid = (L+R)/2;
            ans = Judge(mid);
            if(ans==0)
                L = R = mid;
            if(ans>0)
                L = mid + 1;
            if(ans<0)
                R = mid - 1;
        }

        printf("%I64d\n", L);
    }

    return 0;
}