/*
JOI 2022 Problem 2
Solution:
First, set A_i = max(A_i, B_i) to avoid edge cases.
Then, binary search on the answer (makes the problem easier).
Start with Bitaro attending every class.
Then, for each course in which he overprepares, calculate the # of classes that can be skipped.
See if those missed courses (in total) can make up the ones he underprepares for.
*/
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define sz(x) ((int) x.size())
#define all(x) x.begin(), x.end()
#define debug if (true) cerr
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
bool check(int N, ll M, vi& A, vi& B, ll K) {
vector<ll> prepLeft(N, K);
// Initially, go to all the classes
rep(i, 0, N) prepLeft[i] -= M * A[i];
// Count number of free classes (to be reassigned)
// Also count number of needed self study periods
ll numFree = 0;
ll numSelf = 0;
rep(i, 0, N) {
if (prepLeft[i] <= -A[i]) {
// Overprepared
numFree += min(-prepLeft[i] / A[i], M);
} else if (prepLeft[i] > 0) {
// Underprepared
numSelf += (prepLeft[i]+B[i]-1) / B[i];
}
}
// Possible if enough free periods exist
return numFree >= numSelf;
}
void solve() {
int N; cin >> N;
ll M; cin >> M;
vi A(N), B(N);
rep(i, 0, N) cin >> A[i];
rep(i, 0, N) {
cin >> B[i];
A[i] = max(A[i], B[i]);
}
// Binary search on the answer
ll ansLow = 0, ansHigh = 1e18;
while (ansLow < ansHigh) {
ll mid = (ansLow+ansHigh+1) / 2;
if (check(N, M, A, B, mid)) ansLow = mid;
else ansHigh = mid - 1;
}
ll ans = ansLow;
cout << ans << '\n';
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin.exceptions(cin.failbit);
solve();
return 0;
}
