#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
#define rep(i,n) for(int (i) = 0; i<(n); i++)

const int N = 100000;
const ll inf = 1e18;

int main(){
    int n; ll k;
    cin >> n >> k;
    vector<ll> a(n), b(n);
    rep(i,n) cin >> a[i];
    rep(i,n) cin >> b[i];

    // 制約
    assert(1<=n && n<=N);
    assert(1<=k && k<=inf);
    rep(i,n) assert(1<=a[i] && a[i]<=N);
    rep(i,n) assert(1<=b[i] && b[i]<=N);

    // div[i] : i の約数
    vector<vector<int>> div(N+10);
    for(int i=1; i<=N+5; i++){
        int j = i;
        while(j<=N+5){
            div[j].push_back(i);
            j += i;
        }
    }

    // gcd を div の値にするのに必要な操作回数
    vector<set<P>> st(n);
    rep(i,n){
        ll min_cost = inf;
        for(int j = div[a[i]].size()-1; 0<=j; j--){
            ll p = div[a[i]][j];
            if(b[i]%p){
                ll cost = p-b[i]%p;
                if(cost<min_cost){
                    st[i].insert(P(p,cost));
                    min_cost = cost;
                }
            }
        }
    }

    // 二分探索
    int ok = 1, ng = *min_element(a.begin(),a.end())+1;
    while(1<abs(ok-ng)){
        int mid = (ok+ng)/2;
        ll sum = 0; // 加算が必要な回数
        rep(i,n) sum += (*st[i].lower_bound(P(mid,0))).second;
        if(sum<=k) ok = mid;
        else ng = mid;
    }

    cout << ok << endl;

}