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

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

题目的大致意思是有k克的魔法面粉，可以变成任意一种面粉。要做一个饼干，需要n中面粉。第二行是做一个饼干所需要的第i种面粉的量。第三行是第i种面粉的含量。求最多能做多少个饼干。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

long long n, k, a[100010], b[100010];
bool judge(long long x) {
    long long res = k;
    for (long long i = 0; i < n; i++) {
        if (x > b[i]/a[i] ) {
            if (res < a[i]*x - b[i]) return false;
            else res -= a[i]*x - b[i];
        }
    }
    return true;
}
int main() {
    while (scanf("%lld%lld", &n, &k) != EOF) {
        for (long long i = 0; i < n; i++) {
            scanf("%lld", &a[i]);
        }
        for (long long j = 0; j < n; j++) {
            scanf("%lld", &b[j]);
        }
        long long lb = 0, ub = 2e9 + 10;
        long long ans = 0;
        while (ub >= lb) {
            long long mid = (lb + ub)/2;
            if (judge(mid)) {
                ans = mid;
                lb = mid + 1;
            }
            else ub = mid - 1;
        }
        printf("%lld
", ans);
    }
    return 0;
}